INFINITE SERIES KEITH CONRAD

Size: px
Start display at page:

Download "INFINITE SERIES KEITH CONRAD"

Transcription

1 INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal limit process: ifiite series. The label series is just aother ame for a sum. A ifiite series is a sum with ifiitely may terms, such as (.) The idea of a ifiite series is familiar from decimal expasios, for istace the expasio ca be writte as π = π = , so π is a ifiite sum of fractios. Decimal expasios like this show that a ifiite series is ot a paradoxical idea, although it may ot be clear how to deal with o-decimal ifiite series like (.) at the momet. Ifiite series provide two coceptual isights ito the ature of the basic fuctios met i high school (ratioal fuctios, trigoometric ad iverse trigoometric fuctios, expoetial ad logarithmic fuctios). First of all, these fuctios ca be expressed i terms of ifiite series, ad i this way all these fuctios ca be approximated by polyomials, which are the simplest kids of fuctios. That simpler fuctios ca be used as approximatios to more complicated fuctios lies behid the method which calculators ad computers use to calculate approximate values of fuctios. The secod isight we will have usig ifiite series is the close relatioship betwee fuctios which seem at first to be quite differet, such as expoetial ad trigoometric fuctios. Two other applicatios we will meet are a proof by calculus that there are ifiitely may primes ad a proof that e is irratioal. 2. Defiitios ad basic examples Before discussig ifiite series we discuss fiite oes. A fiite series is a sum a + a 2 + a a N, where the a i s are real umbers. I terms of Σ-otatio, we write N a + a 2 + a a N = a. =

2 2 KEITH CONRAD For example, (2.) N = = N. The sum i (2.) is called a harmoic sum, for istace = A very importat class of fiite series, more importat tha the harmoic oes, are the geometric series N (2.2) + r + r r N = r. A example is = ( ) = 2 = = 2 = = The geometric series (2.2) ca be summed up exactly, as follows. Theorem 2.. Whe r, the series (2.2) is Proof. Let The + r + r r N = rn+ r S N = + r + + r N. rs N = r + r r N+. = rn+ r. These sums overlap i r + + r N, so subtractig rs N from S N gives ( r)s N = r N+. Whe r we ca divide ad get the idicated formula for S N. Example 2.2. For ay N, Example 2.3. For ay N, N = 2N N = (/2)N+ /2 = 2 N+. = 2 2 N. It is sometimes useful to adjust the idexig o a sum, for istace (2.3) = 2 = = 99 = ( + )2.

3 INFINITE SERIES 3 Comparig the two Σ s i (2.3), otice how the reumberig of the idices affects the expressio beig summed: if we subtract from the bouds of summatio o the Σ the we add to the idex o the iside to maitai the same overall sum. As aother example, = 2 = ( + 3) 2. =3 Wheever the idex is shifted dow (or up) by a certai amout o the Σ it has to be shifted up (or dow) by a compesatig amout iside the expressio beig summed so that we are still addig the same umbers. This is aalogous to the effect of a additive chage of variables i a itegral, e.g., (x + 5) 2 dx = 6 5 = u 2 du = 6 5 x 2 dx, where u = x + 5 (ad du = dx). Whe the variable iside the itegral is chaged by a additive costat, the bouds of itegratio are chaged additively i the opposite directio. Now we defie the meaig of ifiite series, such as (.). The basic idea is that we look at the sum of the first N terms, called a partial sum, ad see what happes i the limit as N. Defiitio 2.4. Let a, a 2, a 3,... be a ifiite sequece of real umbers. a is defied to be N a = lim a. N = The ifiite series If the limit exists i R the we say a is coverget. If the limit does ot exist or is ± the a is called diverget. Notice that we are ot really addig up all the terms i a ifiite series at oce. We oly add up a fiite umber of the terms ad the see how thigs behave i the limit as the (fiite) umber of terms teds to ifiity: a ifiite series is defied to be the limit of its sequece of partial sums. Example 2.5. Usig Example 2.3, So 2 = lim N N = is a coverget series ad its value is = lim N 2 2 N = 2. Buildig o this example we ca compute exactly the value of ay ifiite geometric series. Theorem 2.6. For x R, the (ifiite) geometric series x coverges if x < ad diverges if x. If x < the x = x. Proof. For N, Theorem 2. tells us { N x ( x N+ )/( x), if x, = N +, if x =. =

4 4 KEITH CONRAD Whe x <, x N+ as N, so N x = lim N = x x N+ = lim N x = x. Whe x > the umerator x N does ot coverge as N, so x diverges. (Specifically, the limit is if x > ad the partials sums oscillate betwee values tedig to ad if x <.) What if x =? If x = the the N-th partial sum is N + so the series diverges (to ). If x = the N = ( ) oscillates betwee ad, so agai the series ( ) is diverget. We will see that Theorem 2.6 is the fudametal example of a ifiite series. May importat ifiite series will be aalyzed by comparig them to a geometric series (for a suitable choice of x). If we start summig a geometric series ot at, but at a higher power of x, the we ca still get a simple closed formula for the series, as follows. Corollary 2.7. If x < ad m the x = x m + x m+ + x m+2 + = m Proof. The N-th partial sum (for N m) is xm x. x m + x m+ + + x N = x m ( + x + + x N m ). As N, the sum iside the paretheses becomes the stadard geometric series, whose value is /( x). Multiplyig by x m gives the asserted value. Example 2.8. If x < the x + x 2 + x 3 + = x/( x). A diverget geometric series ca diverge i differet ways: the partial sums may ted to or ted to both ad or oscillate betwee ad. The label diverget series does ot always mea the partial sums ted to. All diverget meas is ot coverget. Of course i a particular case we may kow the partial sums do ted to, ad the we would say the series diverges to. The covergece of a series is determied by the behavior of the terms a for large. If we chage (or omit) ay iitial set of terms i a series we do ot chage the covergece of divergece of the series. Here is the most basic geeral feature of coverget ifiite series. Theorem 2.9. If the series a coverges the a as. Proof. Let S N = a + a a N ad let S = a be the limit of the partial sums S N as N. The as N, a N = S N S N S S =. While Theorem 2.9 is formulated for coverget series, its mai importace is as a divergece test : if the geeral term i a ifiite series does ot ted to the the series diverges. For example, Theorem 2.9 gives aother reaso that a geometric series x diverges if x, because i this case x does ot ted to : x = x for all.

5 INFINITE SERIES 5 It is a ufortuate fact of life that the coverse of Theorem 2.9 is geerally false: if a we have o guaratee that a coverges. Here is the stadard illustratio. Example 2.. Cosider the harmoic series turs out that =.. Although the geeral term teds to it To show this we will compare N = / with N dt/t = log N. Whe t, /t /. Itegratig this iequality from to + gives + dt/t /, so N = N = + dt t = N+ dt t = log(n + ). Therefore the N-th partial sum of the harmoic series is bouded below by log(n + ). Sice log(n + ) as N, so does the N-th harmoic sum, so the harmoic series diverges. We will see later that the lower boud log(n + ) for the N-th harmoic sum is rather sharp, so the harmoic series diverges slowly sice the logarithm fuctio diverges slowly. The divergece of the harmoic series is ot just a couterexample to the coverse of Theorem 2.9, but ca be exploited i other cotexts. Let s use it to prove somethig about prime umbers! Theorem 2.. There are ifiitely may prime umbers. Proof. (Euler) We will argue by cotradictio: assumig there are oly fiitely may prime umbers we will cotradict the divergece of the harmoic series. Cosider the product of /( /p) as p rus through all prime umbers. Sums are deoted with a Σ ad products are deoted with a Π (capital pi), so our product is writte as p /p = /2 /3 /5. Sice we assume there are fiitely may primes, this is a fiite product. Now expad each factor ito a geometric series: /p = ( + p + p 2 + p ) 3 +. p p Each geometric series is greater tha ay of its partial sums. Pick ay iteger N 2 ad trucate each geometric series at the N-th term, givig the iequality (2.4) /p > ( + p + p 2 + p ) p N. p p O the right side of (2.4) we have a fiite product of fiite sums, so we ca compute this usig the distributive law ( super FOIL i the termiology of high school algebra). That is, take oe term from each factor, multiply them together, ad add up all these products. What umbers do we get i this way o the right side of (2.4)? A product of reciprocal itegers is a reciprocal iteger, so we obtai a sum of / as rus over certai positive itegers. Specifically, the s we meet are those whose prime factorizatio does t

6 6 KEITH CONRAD ivolve ay prime with expoet greater tha N. Ideed, for ay positive iteger >, if we factor ito primes as = p e pe 2 2 per r the e i < for all i. (Sice p i 2, 2 e i, ad 2 m > m for ay m so > e i.) Thus if N ay of the prime factors of appear i with expoet less tha N, so / is a umber which shows up whe multiplyig out (2.4). Here we eed to kow we are workig with a product over all primes. Sice / occurs i the expasio of the right side of (2.4) for N ad all terms i the expasio are positive, we have the lower boud estimate (2.5) p N /p >. Sice the left side is idepedet of N while the right side teds to as N, this iequality is a cotradictio so there must be ifiitely may primes. We ed this sectio with some algebraic properties of coverget series: termwise additio ad scalig. Theorem 2.2. If a ad b coverge the (a + b ) coverges ad + b ) = (a a + b. If a coverges the for ay umber c the series ca coverges ad a. ca = c Proof. Let S N = N = a ad T N = N = b. Set S = a ad T = b, so S N S ad T N T as N. The by kow properties of limits of sequeces, S N + T N S + T ad cs N cs as N. These are the coclusios of the theorem. Example 2.3. We have ad ( 2 + ) 3 = = /2 + /3 = = 5 4 = 5 /4 = Positive series I this sectio we describe several ways to show that a ifiite series a coverges whe a > for all. There is somethig special about series where all the terms are positive. What is it? The sequece of partial sums is icreasig: S N+ = S N + a N, so S N+ > S N for every N. A icreasig sequece has two possible limitig behaviors: it coverges or it teds to. (That is, divergece ca oly happe i oe way: the partial sums ted to.) We ca tell if a icreasig sequece coverges by showig it is bouded above. The it coverges ad its limit is a upper boud o the whole sequece. If the terms i a icreasig sequece are ot bouded above the the sequece

7 INFINITE SERIES 7 is ubouded ad teds to. Compare this to the sequece of partial sums N = ( ), which are bouded but do ot coverge (they oscillate betwee ad ). Let s summarize this property of positive series. If a > for every ad there is a costat C such that all partial sums satisfy N = a < C the a coverges ad a C. O the other had, if there is o upper boud o the partial sums the a =. We will use this over ad over agai to explai various covergece tests for positive series. Theorem 3. (Itegral Test). Suppose a = f() where f is a cotiuous decreasig positive fuctio. The a coverges if ad oly if f(x) dx coverges. Proof. We will compare the partial sum N = a ad the partial itegral N f(x) dx. Sice f(x) is a decreasig fuctio, x + = a + = f( + ) f(x) f() = a. Itegratig these iequalities betwee ad + gives (3.) a + + sice the iterval [, + ] has legth. Therefore (3.2) ad (3.3) N a = N a = a + = Combiiig (3.2) ad (3.3), (3.4) N = + N a a + =2 N+ f(x) dx N =2 N = f(x) dx a. f(x) dx = N+ f(x) dx N f(x) dx = f() + f(x) dx. N a f() + f(x) dx. Assume that f(x) dx coverges. Sice f(x) is a positive fuctio, N f(x) dx < f(x) dx. Thus by the secod iequality i (3.4) N a < f() + = f(x) dx, so the partial sums are bouded above. Therefore the series a coverges. Coversely, assume a coverges. The the partial sums N = a are bouded above (by the whole series a ). Every partial itegral of f(x) dx is bouded above by oe of these partial sums accordig to the first iequality of (3.4), so the partial itegrals are bouded above ad thus the improper itegral f(x) dx coverges. Example 3.2. We were implicitly usig the itegral test with f(x) = /x whe we proved the harmoic series diverged i Example 2.. For the harmoic series, (3.4) becomes (3.5) log(n + ) N = + log N.

8 8 KEITH CONRAD Thus the harmoic series diverges logarithmically. The followig table illustrates these bouds whe N is a power of. N log(n + ) N = / + log N Table. Example 3.3. Geeralizig the harmoic series, cosider p = + 2 p + 3 p + 4 p + where p >. This is called a p-series. Its covergece is related to the itegral dx/x p. Sice the itegrad /x p has a ati-derivative that ca be computed explicitly, we ca compute the improper itegral: { dx x p = p, if p >,, if < p. Therefore /p coverges whe p > ad diverges whe < p. For example, /2 coverges ad / diverges. Example 3.4. The series 2 log 2 diverges by the itegral test sice dx x log x = log log x 2 =. (We stated the itegral test with a lower boud of but it obviously adapts to series where the first idex of summatio is somethig other tha, as i this example.) Example 3.5. The series, with terms just slightly smaller tha i the previous (log ) 2 2 example, coverges. Usig the itegral test, dx x(log x) = log x = log 2, which is fiite. 2 Because the covergece of a positive series is itimately tied up with the boudedess of its partial sums, we ca determie the covergece or divergece of oe positive series from aother if we have a iequality i oe directio betwee the terms of the two series. This leads to the followig covergece test. Theorem 3.6 (Compariso test). Let < a b for all. If b coverges the a coverges. If a diverges the b diverges. 2

9 Proof. Sice < a b we have (3.6) INFINITE SERIES 9 N a = ad both sequeces of partial sums are icreasig or stay the same (they are odecreasig ). If b coverges the the partial sums o the right side of (3.6) are bouded, so the partial sums o the left side are bouded, ad therefore a coverges because a icreasig (really, a odecreasig) bouded sequece of real umbers coverges. If a diverges the N = a (see the secod paragraph of this sectio agai), so N = b, so b diverges. Example 3.7. If x < the x / coverges sice x / x so the series is bouded above by the geometric series x, which coverges. Notice that, ulike the geometric series, we are ot givig ay kid of closed form expressio for x /. All we are sayig is that the series coverges. The series x / does ot coverge for x : if x = it is the harmoic series ad if x > the the geeral term x / teds to (ot ) so the series diverges to. Example 3.8. The itegral test is a special case of the compariso test, where we compare a with + f(x) dx ad with f(x) dx. Just as covergece of a series does ot deped o the behavior of the iitial terms of the series, what is importat i the compariso test is a iequality a b for large. If this iequality happes ot to hold for small but does hold for all large the the compariso test still works. The followig two examples use this exteded compariso test whe the form of the compariso test i Theorem 3.6 does ot strictly apply (because the iequality a b does t hold for small ). Example 3.9. If x < the x coverges. This is obvious for x =. We ca t get covergece for < x < by a compariso with x because the iequality goes the wrog way: x > x, so covergece of x does t help us show covergece of x. However, let s write x = x /2 x /2. Sice < x <, x /2 as, so x /2 for large (depedig o the specific value of x, e.g, for x = /2 the iequality is false at = 3 but is correct for 4). Thus for all large we have x x /2 so x coverges by compariso with the geometric series x/2 = ( x). We have ot obtaied ay kid of closed formula for x, however. If x the x diverges sice x for all, so the geeral term does ot ted to. Example 3.. If x the the series x /! coverges. This is obvious for x =. For x > we will use the compariso test ad the lower boud estimate! > /e. Let s recall that this lower boud o! comes from Euler s itegral formula The for positive x (3.7)! = t e t dt > N = b t e t dt > x! < x ( ex ) /e =. e t dt = e.

10 KEITH CONRAD Here x is fixed ad we should thik about what happes as grows. The ratio ex/ teds to as, so for large we have ex (3.8) < 2 (ay positive umber less tha will do here; we just use /2 for cocreteess). Raisig both sides of (3.8) to the -th power, ( ex ) < 2. Comparig this to (3.7) gives x /! < /2 for large, so the terms of the series x /! drop off faster tha the terms of the coverget geometric series /2 whe we go out far eough. The exteded compariso test ow implies covergece of x /! if x >. Example 3.. The series Example coverges sice < 5 2 ad coverges by 52 Let s try to apply the compariso test to /(52 ). The rate of growth of the th term is like /(5 2 ), but the iequality ow goes the wrog way: 5 2 < 5 2 for all. So we ca t quite use the compariso test to show /(52 ) coverges (which it does). The followig limitig form of the compariso test will help here. Theorem 3.2 (Limit compariso test). Let a, b > ad suppose the ratio a /b has a positive limit. The a coverges if ad oly if b coverges. Proof. Let L = lim a /b. For all large, so L 2 < a b < 2L, L (3.9) 2 b < a < 2Lb for large. Sice L/2 ad 2L are positive, the covergece of b is the same as covergece of (L/2)b ad (2L)b. Therefore the first iequality i (3.9) ad the compariso test show covergece of a implies covergece of b. The secod iequality i (3.9) ad the compariso test show covergece of b implies covergece of a. (Although (3.9) may ot be true for all, it is true for large, ad that suffices to apply the exteded compariso test.) The way to use the limit compariso test is to replace terms i a series by simpler terms which grow at the same rate (at least up to a scalig factor). The aalyze the series with those simpler terms. Example 3.3. We retur to 5 2. As, the th term grows like /52 : /(5 2 ) /5 2.

11 INFINITE SERIES Sice /(52 ) coverges, so does /(52 ). I fact, if a is ay sequece whose growth is like / 2 up to a scalig factor the a coverges. I particular, this is a simpler way to settle the covergece i Example 3. tha by usig the iequalities i Example 3.. Example 3.4. To determie if coverges we replace the th term with the expressio = 2 3/2, which grows at the same rate. The series /(23/2 ) coverges, so the origial series coverges. The limit compariso test uderlies a importat poit: the covergece of a series a is ot assured just by kowig if a, but it is assured if a rapidly eough. That is, the rate at which a teds to is ofte (but ot always) a meas of explaiig the covergece of a. The differece betwee someoe who has a ituitive feelig for covergece of series ad someoe who ca oly determie covergece o a case-by-case basis usig memorized rules i a mechaical way is probably idicated by how well the perso uderstads the coectio betwee covergece of a series ad rates of growth (really, decay) of the terms i the series. Armed with the covergece of a few basic series (like geometric series ad p-series), the covergece of most other series ecoutered i a calculus course ca be determied by the limit compariso test if you uderstad well the rates of growth of the stadard fuctios. Oe exceptio is if the series has a log term, i which case it might be useful to apply the itegral test. 4. Series with mixed sigs All our previous covergece tests apply to series with positive terms. They also apply to series whose terms are evetually positive (just omit ay iitial egative terms to get a positive series) or evetually egative (egate the series, which does t chage the covergece status, ad ow we re reduced to a evetually positive series). What do ca we do for series whose terms are ot evetually all positive or evetually all egative? These are the series with ifiitely may positive terms ad ifiitely may egative terms. Example 4.. Cosider the alteratig harmoic series ( ) = The usual harmoic series diverges, but the alteratig sigs might have differet behavior. The followig table collects some of the partial sums. N N = ( ) / Table 2.

12 2 KEITH CONRAD These partial sums are ot clear evidece i favor of covergece: after, terms we oly get apparet stability i the first 2 decimal digits. The harmoic series diverges slowly, so perhaps the alteratig harmoic series diverges too (ad slower). Defiitio 4.2. A ifiite series where the terms alterate i sig is called a alteratig series. A alteratig series startig with a positive term ca be writte as ( ) a where a > for all. If the first term is egative the a reasoable geeral otatio is ( ) a where a > for all. Obviously egatig a alteratig series of oe type turs it ito the other type. Theorem 4.3 (Alteratig series test). A alteratig series whose -th term i absolute value teds mootoically to as is a coverget series. Proof. We will work with alteratig series which have a iitial positive term, so of the form ( ) a. Sice successive partial sums alterate addig ad subtractig a amout which, i absolute value, steadily teds to, the relatio betwee the partial sums is idicated by the followig iequalities: s 2 < s 4 < s 6 < < s 5 < s 3 < s. I particular, the eve-idexed partial sums are a icreasig sequece which is bouded above (by s, say), so the partial sums s 2m coverge. Sice s 2m+ s 2m = a 2m+, the odd-idexed partial sums coverge to the same limit as the eve-idexed partial sums, so the sequece of all partial sums has a sigle limit, which meas ( ) a coverges. Example 4.4. The alteratig harmoic series ( ) coverges. Example 4.5. While the p-series /p coverges oly for p >, the alteratig p-series ( ) / p coverges for the wider rage of expoets p > sice it is a alteratig series satisfyig the hypotheses of the alteratig series test. Example 4.6. We saw x / coverges for x < (compariso to the geometric series) i Example 3.7. If x < the x / coverges by the alteratig series test: the egativity of x makes the terms x / alterate i sig. To apply the alteratig series test we eed x + + < x for all, which is equivalet after some algebra to x < + for all, ad this is true sice x < ( + )/. Example 4.7. I Example 3. the series x /! was see to coverge for x. If x < the series is a alteratig series. Does it fit the hypotheses of the alteratig series test? We eed x + ( + )! < x! for all, which is equivalet to x < +

13 INFINITE SERIES 3 for all. For each particular x < this may ot be true for all (if x = 3 it fails for = ad 2), but it is defiitely true for all sufficietly large. Therefore the terms of x /! are a evetually alteratig series. Sice what matters for covergece is the log-term behavior ad ot the iitial terms, we ca apply the alteratig series test to see x /! coverges by just igorig a iitial piece of the series. Whe a series has ifiitely may positive ad egative terms which are ot strictly alteratig, covergece may ot be easy to check. For example, the trigoometric series si(x) = si x + si(2x) 2 + si(3x) 3 + si(4x) 4 turs out to be coverget for every umber x, but the arragemet of positive ad egative terms i this series ca be quite subtle i terms of x. The proof that this series coverges uses the techique of summatio by parts, which wo t be discussed here. While it is geerally a delicate matter to show a o-alteratig series with mixed sigs coverges, there is oe useful property such series might have which does imply covergece. It is this: if the series with all terms made positive coverges the so does the origial series. Let s give this idea a ame ad the look at some examples. + Defiitio 4.8. A series a is absolutely coverget if a coverges. Do t cofuse a with a ; absolute covergece refers to covergece if we drop the sigs from the terms i the series, ot from the series overall. The differece betwee a ad a is at most a sigle sig, while there is a substatial differece betwee a ad a if the a s have may mixed sigs; that is the differece which absolute covergece ivolves. Example 4.9. We saw x / coverges if x < i Example 3.7. Sice x / = x /, the series x / is absolutely coverget if x <. Note the alteratig harmoic series ( ) / is ot absolutely coverget, however. Example 4.. I Example 3. we saw x /! coverges for all x. Sice x /! = x /!, the series x /! is absolutely coverget for all x. Example 4.. By Example 3.9, x coverges absolutely if x <. Example 4.2. The two series si(x) 2, cos(x) 2, where the th term has deomiator 2, are absolutely coverget for ay x sice the absolute value of the th term is at most / 2 ad /2 coverges. The relevace of absolute covergece is two-fold: ) we ca use the may covergece tests for positive series to determie if a series is absolutely coverget, ad 2) absolute covergece implies ordiary covergece. We just illustrated the first poit several times. Let s show the secod poit. Theorem 4.3 (Absolute covergece test). Every absolutely coverget series is a coverget series.

14 4 KEITH CONRAD Proof. We give two proofs. The first oe, which is simpler, exploits special features of real umbers. The secod proof ca be applied i more geeral settigs that use ifiite series of umbers other tha real umbers (e.g., ifiite series of complex umbers or ifiite series of fuctios). First proof: For all we have a a a, so a + a 2 a. Sice a coverges, so does 2 a (to twice the value). Therefore by the compariso test we obtai covergece of (a + a ). Subtractig a from this shows a coverges. Secod proof: To show a coverges, we wat to show the partial sums s N = a + a a N have a limit. We will show the sequece of partial sums {s N } is a Cauchy sequece: for all ε > there is a positive iteger M such that if N, N M the s N s N < ε. Sice a is assumed to coverge, the partial sums t N = a + a a N form a Cauchy sequece: for all ε > there is a positive iteger M such that if N, N M the t N t N < ε. Takig N N without loss of geerality, we have t N t N = a N a N, so N N M = a N a N < ε. Thus N N M = s N s N = a N a N a N a N = t N t N < ε, so {s N } is a Cauchy sequece ad therefore it coverges i R. We ca ow say the series x / ad x coverge for x < ad x /! coverges for all x by our treatmet of these series for positive x aloe i Sectio 3. The argumet we gave i Examples 4.6 ad 4.7 for the covergece of x / ad x /! whe x <, usig the alteratig series test, ca be avoided. (But we do eed the alteratig series test to show x / coverges at x =.) Corollary 4.4. If a b ad b coverges the a coverges. Proof. By the compariso test (Theorem 3.6), covergece of b implies covergece of a, ad by Theorem 4.3 a coverges. A series which coverges but is ot absolutely coverget is called coditioally coverget. A example of a coditioally coverget series is the alteratig harmoic series ( ) /. The distictio betwee absolutely coverget series ad coditioally coverget series might seem kid of mior, sice it ivolves whether or ot a particular covergece test (the absolute covergece test) works o that series. But the distictio is actually profoud, ad is icely illustrated by the followig example of Dirichlet (837). Example 4.5. Let L = /2 + /3 /4 + /5 /6 + be the alteratig harmoic series. Cosider the rearragemet of the terms where we follow a positive term by two egative terms:

15 INFINITE SERIES 5 If we add each positive term to the egative term followig it, we obtai , which is precisely half the alteratig harmoic series (multiply L by /2 ad see what you get termwise). Therefore, by rearragig the terms of the alteratig harmoic series we obtaied a ew series whose sum is (/2)L istead of L. If (/2)L = L, does t that mea = 2? What happeed? This example shows additio of terms i a ifiite series is ot geerally commutative, ulike with fiite series. I fact, this feature is characteristic of the coditioally coverget series, as see i the followig amazig theorem. Theorem 4.6 (Riema, 854). If a ifiite series is coditioally coverget the it ca be rearraged to sum up to ay desired value. If a ifiite series is absolutely coverget the all of its rearragemets coverge to the same sum. For istace, the alteratig harmoic series ca be rearraged to sum up to, to 5.673, to π, or to ay other umber you wish. That we rearraged it i Example 4.5 to sum up to half its usual value was special oly i the sese that we could make the rearragemet achievig that effect quite explicit. Theorem 4.6, whose proof we omit, is called Riema s rearragemet theorem. It idicates that absolutely coverget series are better behaved tha coditioally coverget oes. The ordiary rules of algebra for fiite series geerally exted to absolutely coverget series but ot to coditioally coverget series. 5. Power series The simplest fuctios are polyomials: c + c x + + c d x d. I differetial calculus we lear how to locally approximate may fuctios by liear fuctios (the taget lie approximatio). We ow itroduce a idea which will let us give a exact represetatio of fuctios i terms of polyomials of ifiite degree. The precise ame for this idea is a power series. Defiitio 5.. A power series is a fuctio of the form f(x) = c x. For istace, a polyomial is a power series where c = for large. The geometric series x is a power series. It oly coverges for x <. We have also met other power series, like x, x x,!. These three series coverge o the respective itervals [, ), (, ) ad (, ). For ay power series c x it is a basic task to fid those x where the series coverges. It turs out i geeral, as i all of our examples, that a power series coverges o a iterval. Let s see why. Theorem 5.2. If a power series c x coverges at x = b the it coverges absolutely for x < b. Proof. Sice c b coverges the geeral term teds to, so c b for large. If x < b the c x = c b x, b

16 6 KEITH CONRAD which is at most x/b for large. The series x/b coverges sice it s a geometric series ad x/b <. Therefore by the (exteded) compariso test c x coverges, so c x is absolutely coverget. Example 5.3. If a power series coverges at 7 the it coverges o the iterval ( 7, 7). We ca t say for sure how it coverges at 7. Example 5.4. If a power series diverges at 7 the it diverges for x > 7: if it were to coverge at a umber x where x > 7 the it would coverge i the iterval ( x, x ) so i particular at 7, a cotradictio. What these two examples illustrate, as cosequeces of Theorem 5.2, is that the values of x where a power series c x coverges is a iterval cetered at, so of the form ( r, r), [ r, r), ( r, r], [ r, r] for some r. We call r the radius of covergece of the power series. The oly differece betwee these differet itervals is the presece or absece of the edpoits. All possible types of iterval of covergece ca occur: x has iterval of covergece (, ), x / has iterval of covergece [, ), ( ) x / has iterval of covergece (, ] ad x / 2 (a ew example we have ot met before) has iterval of covergece [, ]. The radius of covergece ad the iterval of covergece are closely related but should ot be cofused. The iterval is the actual set where the power series coverges. The radius is simply the half-legth of this set (ad does t tell us whether or ot the edpoits are icluded). If we do t care about covergece behavior o the boudary of the iterval of covergece tha we ca get by just kowig the radius of covergece: the series always coverges (absolutely) o the iside of the iterval of covergece. For the practical computatio of the radius of covergece i basic examples it is coveiet to use a ew covergece test for positive series. Theorem 5.5 (Ratio test). If a > for all, assume a + q = lim a exists. If q < the a coverges. If q > the a diverges. If q = the o coclusio ca be draw. Proof. Assume q <. Pick s betwee q ad : q < s <. Sice a + /a q, we have a + /a < s for all large, say for. The a + < sa whe, so a + < sa, a +2 < sa + < s 2 a, a +3 < sa +2 < s 3 a, ad more geerally a +k < s k a for ay k. Writig + k as we have = a < s a = a s s. Therefore whe the umber a is bouded above by a costat multiple of s. Hece a coverges by compariso to a costat multiple of the coverget geometric series s. I the other directio, if q > the pick s with < s < q. A argumet similar to the previous oe shows a grows at least as quickly as a costat multiple of s, but this time s diverges sice s >. So a diverges too. Whe q = we ca t make a defiite coclusio sice both / ad /2 have q =.

17 INFINITE SERIES 7 Example 5.6. We give a proof that x has radius of covergece which is shorter tha Examples 3.9 ad 4.. Take a = x = x. The a + = + x x a as. Thus if x < the series x is absolutely coverget by the ratio test. If x > this series is diverget. Notice the ratio test does ot tell us what happes to x whe x = ; we have to check x = ad x = idividually. Example 5.7. A similar argumet shows x / has radius of covergece sice as. x + /( + ) x / = x x + Example 5.8. We show the series x /! coverges absolutely for all x. Usig the ratio test we look at x + /( + )! x = x /! + as. Sice this limit is less tha for all x, we are doe by the ratio test. The radius of covergece is ifiite. Power series are importat for two reasos: they give us much greater flexibility to defie ew kids of fuctios ad may stadard fuctios ca be expressed i terms of a power series. Ituitively, if a kow fuctio f(x) has a power series represetatio o some iterval aroud, say f(x) = c x = c + c x + c 2 x 2 + c 3 x 3 + c 4 x 4 + for x < r, the we ca guess a formula for c i terms of the behavior of f(x). To begi we have f() = c. Now we thik about power series as somethig like ifiite degree polyomials. We kow how to differetiate a polyomial by differetiatig each power of x separately. Let s assume this works for power series too i the iterval of covergece: so we expect Now let s differetiate agai: f (x) = c x = c + 2c 2 x + 3c 3 x 2 + 4c 4 x 3 +, f () = c. f (x) = 2 ( )c x 2 = 2c 2 + 6c 3 x + 2c 4 x 2 +, so f () = 2c 2. Differetiatig a third time (formally) ad settig x = gives The geeral rule appears to be f (3) () = 6c 3. f () () =!c,

18 8 KEITH CONRAD so we should have c = f () ().! This procedure really is valid, accordig to the followig theorem whose log proof is omitted. Theorem 5.9. Ay fuctio represeted by a power series i a ope iterval ( r, r) is ifiitely differetiable i ( r, r) ad its derivatives ca be computed by termwise differetiatio of the power series. This meas our previous calculatios are justified: if a fuctio f(x) ca be writte i the form c x i a iterval aroud the we must have (5.) c = f () ().! I particular, a fuctio has at most oe expressio as a power series c x aroud. Ad a fuctio which is ot ifiitely differetiable aroud will defiitely ot have a power series represetatio c x. For istace, x has o power series represetatio of this form sice it is ot differetiable at. Remark 5.. We ca also cosider power series f(x) = c (x a), whose iterval of covergece is cetered at a. I this case the coefficiets are give by the formula c = f () (a)/!. For simplicity we will focus o power series cetered at oly. Remark 5.. Eve if a power series coverges o a closed iterval [ r, r], the power series for its derivative eed ot coverge at the edpoits. Cosider f(x) = ( ) x 2 /. It coverges for x but the derivative series is 2( ) x 2, which coverges for x <. Remark 5.2. Because a power series ca be differetiated (iside its iterval of covergece) termwise, we ca discover solutios to a differetial equatio by isertig a power series with ukow coefficiets ito the differetial equatio to get relatios betwee the coefficiets. The a few iitial coefficiets, oce chose, should determie all the higher oes. After we compute the radius of covergece of this ew power series we will have foud a solutio to the differetial equatio i a specific iterval. This idea goes back to Newto. It does ot ecessarily provide us with all solutios to a differetial equatio, but it is oe of the stadard methods to fid some solutios. Example 5.3. If x < the x =. Differetiatig both sides, for x < we have x x = ( x) 2. Multiplyig by x gives us x = x ( x) 2, so we have fially foud a closed form expressio for a ifiite series we first met back i Example 3.9.

19 INFINITE SERIES 9 Example 5.4. If there is a power series represetatio c x for e x the (5.) shows c = /!. The oly possible way to write e x as a power series is x! = + x + x2 2 + x3 3! + x4 4! +. Is this really a valid formula for e x? Well, we checked before that this power series coverges for all x. Moreover, calculatig the derivative of this series reproduces the series agai, so this series is a fuctio satisfyig y (x) = y(x). The oly solutios to this differetial equatio are Ce x ad we ca recover C by settig x =. Sice the power series has costat term (so its value at x = is ), its C is, so this power series is e x : Loosely speakig, this meas the polyomials e x =, + x, + x + x2 2, + x + x2 2 + x3 6,... are good approximatios to e x. (Where they are good will deped o how far x is from.) I particular, settig x = gives a ifiite series represetatio of the umber e: (5.2) e = x!.! = ! + 4! + Formula (5.2) ca be used to verify a iterestig fact about e. Theorem 5.5. The umber e is irratioal. Proof. (Fourier) For ay, ( e = + 2! + 3! + + )! ( = + 2! + 3! + +! ( + ) +! ) ( + 2)! + ( + )! + ( + + ( + 2)( + ) + The secod term i paretheses is positive ad bouded above by the geometric series Therefore + + ( + ) 2 + ( + ) 3 + =. ( < e + 2! + 3! + + )!!. Write the sum + /2! + + /! as a fractio with commo deomiator!, say as p /!. Clear the deomiator! to get (5.3) <!e p. So far everythig we have doe ivolves o uproved assumptios. Now we itroduce the ratioality assumptio. If e is ratioal, the!e is a iteger whe is large (sice ay iteger is a factor by! for large ). But that makes!e p a iteger located i the ope iterval (, ), which is absurd. We have a cotradictio, so e is irratioal. ).

20 2 KEITH CONRAD We retur to the geeral task of represetig fuctios by their power series. Eve whe a fuctio is ifiitely differetiable for all x, its power series could have a fiite radius of covergece. Example 5.6. Viewig /( + x 2 ) as /( ( x 2 )) we have the geometric series formula + x 2 = ( x 2 ) = ( ) x 2 whe x 2 <, or equivaletly x <. The series has a fiite iterval of covergece (, ) eve though the fuctio /( + x 2 ) has o bad behavior at the edpoits: it is ifiitely differetiable at every real umber x. Example 5.7. Let f(x) = x /. This coverges for x <. For x < we have f (x) = x = x = x. Whe x <, a atiderivative of /( x) is log( x). Sice f(x) ad log( x) have the same value (zero) at x = they are equal: log( x) = for x <. Replacig x with x ad egatig gives (5.4) log( + x) = x ( ) for x <. The right side coverges at x = (alteratig series) ad diverges at x =. It seems plausible, sice the series equals log( + x) o (, ), that this should remai true at the boudary: does ( ) / equal log 2? (Notice this is the alteratig harmoic series.) We will retur to this later. I practice, if we wat to show a ifiitely differetiable fuctio equals its associated power series (f () ()/!)x o some iterval aroud we eed some kid of error estimate for the differece betwee f(x) ad the polyomial N = (f () ()/!)x. Whe the estimate goes to as N we will have proved f(x) is equal to its power series. Theorem 5.8. If f(x) is ifiitely differetiable the for all N where f(x) = R N (x) = N! Proof. Whe N = the desired result says N = x x f () () x + R N (x),! f(x) = f() + f (N+) (t)(x t) N dt. x This is precisely the fudametal theorem of calculus! f (t) dt.

21 INFINITE SERIES 2 We obtai the N = case from this by itegratio by parts. Set u = f (t) ad dv = dt. The du = f (t) dt ad we (cleverly!) take v = t x (rather tha just t). The x f (t) dt = f t=x x (t)(t x) f (t)(t x) dt so = f ()x + f(x) = f() + f ()x + t= x f (t)(x t) dt, x f (t)(x t) dt. Now apply itegratio by parts to this ew itegral with u = f (t) ad dv = (x t) dt. The du = f (3) (t) dt ad (cleverly) use v = (/2)(x t) 2. The result is so x f (t)(x t) dt = f () x x x f (3) (t)(x t) 2 dt, f(x) = f() + f ()x + f () x 2 + f (3) (t)(x t) 2 dt. 2! 2 Itegratig by parts several more times gives the desired result. We call R N (x) a remaider term. Corollary 5.9. If f is ifiitely differetiable the the followig coditios at the umber x are equivalet: f(x) = (f () ()/!)x, R N (x) as N. Proof. The differece betwee f(x) ad the Nth partial sum of its power series is R N (x), so f(x) equals its power series precisely whe R N (x) as N. Example 5.2. I (5.4) we showed log( + x) equals ( ) x / for x <. What about at x =? For this purpose we wat to show R N () as N whe f(x) = log( + x). To estimate R N () we eed to compute the higher derivatives of f(x): f (x) = + x, f (x) = ( + x) 2, f (3) 2 (x) = ( + x) 3, f (4) 6 (x) = ( + x) 4, ad i geeral for. Therefore so R N () = N! f () (x) = ( ) ( )! ( + x) ( ) N N! ( + t) N+ ( t)n dt = ( ) N ( t) N dt, ( + t) N+ R N () dt ( + t) N+ = N N2 N as N. Sice the remaider term teds to, log 2 = ( ) /.

22 22 KEITH CONRAD Example 5.2. Let f(x) = si x. Its power series is ( ) x2+ (2 + )! = x x3 3! + x5 5! x7 7! +, which coverges for all x by the ratio test. We will show si x equals its power series for all x. This is obvious if x =. Sice si x has derivatives ± cos x ad ± si x, which are bouded by, for x > R N (x) = x N! f (N+) (t)(x t) N dt N! N! x x = xn+ x t N dt x N dt. N! This teds to as N, so si x does equal its power series for x >. If x < the we ca similarly estimate R N (x) ad show it teds to as N, but we ca also use a little trick because we kow si x is a odd fuctio: if x < the x > so si x = si( x) = ( ) ( x)2+ (2 + )! from the power series represetatio of the sie fuctio at positive umbers. Sice ( x) 2+ = x 2+, si x = (2 + )! = ( ) (2 + )!, which shows si x equals its power series for x < too. ( ) x2+ x2+ Example By similar work we ca show cos x equals its power series represetatio everywhere: cos x = ( ) x2 (2)! = x2 2! + x4 4! x6 6! +. The power series for si x ad cos x look like the odd ad eve degree terms i the power series for e x. Is the expoetial fuctio related to the trigoometric fuctios? This idea is suggested because of their power series. Sice e x has a series without the ( ) factors, we ca get a match betwee these series by workig ot with e x but with e ix, where i is a square root of. We have ot defied ifiite series (or eve the expoetial fuctio) usig complex umbers. However, assumig we could make sese of this the we would expect to have e ix = (ix)! = + ix x2 2! ix3 = x2 2! + x4 4! + i = cos x + i si x. 3! + x4 4! + ix5 5! + (x x3 3! + x5 5! )

23 For example, settig x = π would say e iπ =. Replacig x with x i the formula for e ix gives INFINITE SERIES 23 e ix = cos x i si x. Addig ad subtractig the formulas for e ix ad e ix lets us express the real trigoometric fuctios si x ad cos x i terms of (complex) expoetial fuctios: cos x = eix + e ix, si x = eix e ix. 2 2i Evidetly a deeper study of ifiite series should make systematic use of complex umbers! The followig questio is atural: who eeds all the remaider estimate busiess from Theorem 5.8 ad Corollary 5.9? After all, why ot just compute the power series for f(x) ad fid its radius of covergece? That has to be where f(x) equals its power series. Alas, this is false. Example Let f(x) = { e /x2, if x,, if x =. It ca be show that f(x) is ifiitely differetiable at x = ad f () () = for all, so the power series for f(x) has every coefficiet equal to, which meas the power series is the zero fuctio. But f(x) if x, so f(x) is ot equal to its power series aywhere except at x =. Example 5.23 shows that if a fuctio f(x) is ifiitely differetiable for all x ad the power series (f () ()/!)x coverges for all x, f(x) eed ot equal its power series aywhere except at x = (where they must agree sice the costat term of the power series is f()). This is why there is o-trivial cotet i sayig a fuctio ca be represeted by its power series o some iterval. The eed for remaider estimates i geeral is importat.

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

CHAPTER 10 INFINITE SEQUENCES AND SERIES

CHAPTER 10 INFINITE SEQUENCES AND SERIES CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece

More information

MAT1026 Calculus II Basic Convergence Tests for Series

MAT1026 Calculus II Basic Convergence Tests for Series MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you

More information

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special

More information

Chapter 10: Power Series

Chapter 10: Power Series Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because

More information

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet

More information

ENGI Series Page 6-01

ENGI Series Page 6-01 ENGI 3425 6 Series Page 6-01 6. Series Cotets: 6.01 Sequeces; geeral term, limits, covergece 6.02 Series; summatio otatio, covergece, divergece test 6.03 Stadard Series; telescopig series, geometric series,

More information

Part I: Covers Sequence through Series Comparison Tests

Part I: Covers Sequence through Series Comparison Tests Part I: Covers Sequece through Series Compariso Tests. Give a example of each of the followig: (a) A geometric sequece: (b) A alteratig sequece: (c) A sequece that is bouded, but ot coverget: (d) A sequece

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

Sequences. Notation. Convergence of a Sequence

Sequences. Notation. Convergence of a Sequence Sequeces A sequece is essetially just a list. Defiitio (Sequece of Real Numbers). A sequece of real umbers is a fuctio Z (, ) R for some real umber. Do t let the descriptio of the domai cofuse you; it

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information

Ma 530 Infinite Series I

Ma 530 Infinite Series I Ma 50 Ifiite Series I Please ote that i additio to the material below this lecture icorporated material from the Visual Calculus web site. The material o sequeces is at Visual Sequeces. (To use this li

More information

Once we have a sequence of numbers, the next thing to do is to sum them up. Given a sequence (a n ) n=1

Once we have a sequence of numbers, the next thing to do is to sum them up. Given a sequence (a n ) n=1 . Ifiite Series Oce we have a sequece of umbers, the ext thig to do is to sum them up. Give a sequece a be a sequece: ca we give a sesible meaig to the followig expressio? a = a a a a While summig ifiitely

More information

Math 2784 (or 2794W) University of Connecticut

Math 2784 (or 2794W) University of Connecticut ORDERS OF GROWTH PAT SMITH Math 2784 (or 2794W) Uiversity of Coecticut Date: Mar. 2, 22. ORDERS OF GROWTH. Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really

More information

Series III. Chapter Alternating Series

Series III. Chapter Alternating Series Chapter 9 Series III With the exceptio of the Null Sequece Test, all the tests for series covergece ad divergece that we have cosidered so far have dealt oly with series of oegative terms. Series with

More information

Chapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics:

Chapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics: Chapter 6 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals (which is what most studets

More information

Chapter 6: Numerical Series

Chapter 6: Numerical Series Chapter 6: Numerical Series 327 Chapter 6 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals

More information

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe

More information

Roberto s Notes on Series Chapter 2: Convergence tests Section 7. Alternating series

Roberto s Notes on Series Chapter 2: Convergence tests Section 7. Alternating series Roberto s Notes o Series Chapter 2: Covergece tests Sectio 7 Alteratig series What you eed to kow already: All basic covergece tests for evetually positive series. What you ca lear here: A test for series

More information

INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES 11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS

More information

Chapter 7: Numerical Series

Chapter 7: Numerical Series Chapter 7: Numerical Series Chapter 7 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals

More information

Carleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of true-false questions, like the one below.

Carleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of true-false questions, like the one below. Carleto College, Witer 207 Math 2, Practice Fial Prof. Joes Note: the exam will have a sectio of true-false questios, like the oe below.. True or False. Briefly explai your aswer. A icorrectly justified

More information

Advanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology

Advanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology Advaced Aalysis Mi Ya Departmet of Mathematics Hog Kog Uiversity of Sciece ad Techology September 3, 009 Cotets Limit ad Cotiuity 7 Limit of Sequece 8 Defiitio 8 Property 3 3 Ifiity ad Ifiitesimal 8 4

More information

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3 Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x

More information

INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This

More information

MA131 - Analysis 1. Workbook 7 Series I

MA131 - Analysis 1. Workbook 7 Series I MA3 - Aalysis Workbook 7 Series I Autum 008 Cotets 4 Series 4. Defiitios............................... 4. Geometric Series........................... 4 4.3 The Harmoic Series.........................

More information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

Math 113 Exam 4 Practice

Math 113 Exam 4 Practice Math Exam 4 Practice Exam 4 will cover.-.. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for

More information

7 Sequences of real numbers

7 Sequences of real numbers 40 7 Sequeces of real umbers 7. Defiitios ad examples Defiitio 7... A sequece of real umbers is a real fuctio whose domai is the set N of atural umbers. Let s : N R be a sequece. The the values of s are

More information

MA131 - Analysis 1. Workbook 3 Sequences II

MA131 - Analysis 1. Workbook 3 Sequences II MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................

More information

MA131 - Analysis 1. Workbook 9 Series III

MA131 - Analysis 1. Workbook 9 Series III MA3 - Aalysis Workbook 9 Series III Autum 004 Cotets 4.4 Series with Positive ad Negative Terms.............. 4.5 Alteratig Series.......................... 4.6 Geeral Series.............................

More information

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n. 0_0905.qxd //0 :7 PM Page SECTION 9.5 Alteratig Series Sectio 9.5 Alteratig Series Use the Alteratig Series Test to determie whether a ifiite series coverges. Use the Alteratig Series Remaider to approximate

More information

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at

More information

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet

More information

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)

PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e) Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages

More information

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as

More information

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1. SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad

More information

Fall 2013 MTH431/531 Real analysis Section Notes

Fall 2013 MTH431/531 Real analysis Section Notes Fall 013 MTH431/531 Real aalysis Sectio 8.1-8. Notes Yi Su 013.11.1 1. Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters

More information

Additional Notes on Power Series

Additional Notes on Power Series Additioal Notes o Power Series Mauela Girotti MATH 37-0 Advaced Calculus of oe variable Cotets Quick recall 2 Abel s Theorem 2 3 Differetiatio ad Itegratio of Power series 4 Quick recall We recall here

More information

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this

More information

ENGI Series Page 5-01

ENGI Series Page 5-01 ENGI 344 5 Series Page 5-01 5. Series Cotets: 5.01 Sequeces; geeral term, limits, covergece 5.0 Series; summatio otatio, covergece, divergece test 5.03 Series; telescopig series, geometric series, p-series

More information

Practice Test Problems for Test IV, with Solutions

Practice Test Problems for Test IV, with Solutions Practice Test Problems for Test IV, with Solutios Dr. Holmes May, 2008 The exam will cover sectios 8.2 (revisited) to 8.8. The Taylor remaider formula from 8.9 will ot be o this test. The fact that sums,

More information

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals

More information

The Riemann Zeta Function

The Riemann Zeta Function Physics 6A Witer 6 The Riema Zeta Fuctio I this ote, I will sketch some of the mai properties of the Riema zeta fuctio, ζ(x). For x >, we defie ζ(x) =, x >. () x = For x, this sum diverges. However, we

More information

The Growth of Functions. Theoretical Supplement

The Growth of Functions. Theoretical Supplement The Growth of Fuctios Theoretical Supplemet The Triagle Iequality The triagle iequality is a algebraic tool that is ofte useful i maipulatig absolute values of fuctios. The triagle iequality says that

More information

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n = 60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece

More information

MAS111 Convergence and Continuity

MAS111 Convergence and Continuity MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece

More information

Section 11.8: Power Series

Section 11.8: Power Series Sectio 11.8: Power Series 1. Power Series I this sectio, we cosider geeralizig the cocept of a series. Recall that a series is a ifiite sum of umbers a. We ca talk about whether or ot it coverges ad i

More information

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense, 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [

More information

Series Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1

Series Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1 Defiitio: We say that the series S = Series Review i= a i is the sum of the first terms. i= a i coverges if lim S exists ad is fiite, where The above is the defiitio of covergece for series. order to see

More information

Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.

Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book. THE UNIVERSITY OF WARWICK FIRST YEAR EXAMINATION: Jauary 2009 Aalysis I Time Allowed:.5 hours Read carefully the istructios o the aswer book ad make sure that the particulars required are etered o each

More information

10.6 ALTERNATING SERIES

10.6 ALTERNATING SERIES 0.6 Alteratig Series Cotemporary Calculus 0.6 ALTERNATING SERIES I the last two sectios we cosidered tests for the covergece of series whose terms were all positive. I this sectio we examie series whose

More information

SUMMARY OF SEQUENCES AND SERIES

SUMMARY OF SEQUENCES AND SERIES SUMMARY OF SEQUENCES AND SERIES Importat Defiitios, Results ad Theorems for Sequeces ad Series Defiitio. A sequece {a } has a limit L ad we write lim a = L if for every ɛ > 0, there is a correspodig iteger

More information

1 Approximating Integrals using Taylor Polynomials

1 Approximating Integrals using Taylor Polynomials Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................

More information

Math 341 Lecture #31 6.5: Power Series

Math 341 Lecture #31 6.5: Power Series Math 341 Lecture #31 6.5: Power Series We ow tur our attetio to a particular kid of series of fuctios, amely, power series, f(x = a x = a 0 + a 1 x + a 2 x 2 + where a R for all N. I terms of a series

More information

The Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1

The Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1 460_0906.qxd //04 :8 PM Page 69 SECTION 9.6 The Ratio ad Root Tests 69 Sectio 9.6 EXPLORATION Writig a Series Oe of the followig coditios guaratees that a series will diverge, two coditios guaratee that

More information

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =

More information

AP Calculus Chapter 9: Infinite Series

AP Calculus Chapter 9: Infinite Series AP Calculus Chapter 9: Ifiite Series 9. Sequeces a, a 2, a 3, a 4, a 5,... Sequece: A fuctio whose domai is the set of positive itegers = 2 3 4 a = a a 2 a 3 a 4 terms of the sequece Begi with the patter

More information

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will

More information

11.6 Absolute Convergence and the Ratio and Root Tests

11.6 Absolute Convergence and the Ratio and Root Tests .6 Absolute Covergece ad the Ratio ad Root Tests The most commo way to test for covergece is to igore ay positive or egative sigs i a series, ad simply test the correspodig series of positive terms. Does

More information

MA131 - Analysis 1. Workbook 2 Sequences I

MA131 - Analysis 1. Workbook 2 Sequences I MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................

More information

Sequences I. Chapter Introduction

Sequences I. Chapter Introduction Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which

More information

x a x a Lecture 2 Series (See Chapter 1 in Boas)

x a x a Lecture 2 Series (See Chapter 1 in Boas) Lecture Series (See Chapter i Boas) A basic ad very powerful (if pedestria, recall we are lazy AD smart) way to solve ay differetial (or itegral) equatio is via a series expasio of the correspodig solutio

More information

Infinite Series. Definition. An infinite series is an expression of the form. Where the numbers u k are called the terms of the series.

Infinite Series. Definition. An infinite series is an expression of the form. Where the numbers u k are called the terms of the series. Ifiite Series Defiitio. A ifiite series is a expressio of the form uk = u + u + u + + u + () 2 3 k Where the umbers u k are called the terms of the series. Such a expressio is meat to be the result of

More information

Math 116 Second Midterm November 13, 2017

Math 116 Second Midterm November 13, 2017 Math 6 Secod Midterm November 3, 7 EXAM SOLUTIONS. Do ot ope this exam util you are told to do so.. Do ot write your ame aywhere o this exam. 3. This exam has pages icludig this cover. There are problems.

More information

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b) Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig

More information

Testing for Convergence

Testing for Convergence 9.5 Testig for Covergece Remember: The Ratio Test: lim + If a is a series with positive terms ad the: The series coverges if L . The test is icoclusive if L =. a a = L This

More information

( 1) n (4x + 1) n. n=0

( 1) n (4x + 1) n. n=0 Problem 1 (10.6, #). Fid the radius of covergece for the series: ( 1) (4x + 1). For what values of x does the series coverge absolutely, ad for what values of x does the series coverge coditioally? Solutio.

More information

Ma 530 Introduction to Power Series

Ma 530 Introduction to Power Series Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power

More information

Mathematical Methods for Physics and Engineering

Mathematical Methods for Physics and Engineering Mathematical Methods for Physics ad Egieerig Lecture otes Sergei V. Shabaov Departmet of Mathematics, Uiversity of Florida, Gaiesville, FL 326 USA CHAPTER The theory of covergece. Numerical sequeces..

More information

Chapter 4. Fourier Series

Chapter 4. Fourier Series Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,

More information

Math 299 Supplement: Real Analysis Nov 2013

Math 299 Supplement: Real Analysis Nov 2013 Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality

More information

Math 21, Winter 2018 Schaeffer/Solis Stanford University Solutions for 20 series from Lecture 16 notes (Schaeffer)

Math 21, Winter 2018 Schaeffer/Solis Stanford University Solutions for 20 series from Lecture 16 notes (Schaeffer) Math, Witer 08 Schaeffer/Solis Staford Uiversity Solutios for 0 series from Lecture 6 otes (Schaeffer) a. r 4 +3 The series has algebraic terms (polyomials, ratioal fuctios, ad radicals, oly), so the test

More information

Math 19B Final. Study Aid. June 6, 2011

Math 19B Final. Study Aid. June 6, 2011 Math 9B Fial Study Aid Jue 6, 20 Geeral advise. Get plety of practice. There s a lot of material i this sectio - try to do as may examples ad as much of the homework as possible to get some practice. Just

More information

Sequences, Series, and All That

Sequences, Series, and All That Chapter Te Sequeces, Series, ad All That. Itroductio Suppose we wat to compute a approximatio of the umber e by usig the Taylor polyomial p for f ( x) = e x at a =. This polyomial is easily see to be 3

More information

Infinite Series and Improper Integrals

Infinite Series and Improper Integrals 8 Special Fuctios Ifiite Series ad Improper Itegrals Ifiite series are importat i almost all areas of mathematics ad egieerig I additio to umerous other uses, they are used to defie certai fuctios ad to

More information

Chapter 3. Strong convergence. 3.1 Definition of almost sure convergence

Chapter 3. Strong convergence. 3.1 Definition of almost sure convergence Chapter 3 Strog covergece As poited out i the Chapter 2, there are multiple ways to defie the otio of covergece of a sequece of radom variables. That chapter defied covergece i probability, covergece i

More information

TR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT

TR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT TR/46 OCTOBER 974 THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION by A. TALBOT .. Itroductio. A problem i approximatio theory o which I have recetly worked [] required for its solutio a proof that the

More information

MATH 2300 review problems for Exam 2

MATH 2300 review problems for Exam 2 MATH 2300 review problems for Exam 2. A metal plate of costat desity ρ (i gm/cm 2 ) has a shape bouded by the curve y = x, the x-axis, ad the lie x =. (a) Fid the mass of the plate. Iclude uits. Mass =

More information

4.1 Sigma Notation and Riemann Sums

4.1 Sigma Notation and Riemann Sums 0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas

More information

Lesson 10: Limits and Continuity

Lesson 10: Limits and Continuity www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals

More information

Discrete Mathematics for CS Spring 2007 Luca Trevisan Lecture 22

Discrete Mathematics for CS Spring 2007 Luca Trevisan Lecture 22 CS 70 Discrete Mathematics for CS Sprig 2007 Luca Trevisa Lecture 22 Aother Importat Distributio The Geometric Distributio Questio: A biased coi with Heads probability p is tossed repeatedly util the first

More information

MATH4822E FOURIER ANALYSIS AND ITS APPLICATIONS

MATH4822E FOURIER ANALYSIS AND ITS APPLICATIONS MATH48E FOURIER ANALYSIS AND ITS APPLICATIONS 7.. Cesàro summability. 7. Summability methods Arithmetic meas. The followig idea is due to the Italia geometer Eresto Cesàro (859-96). He shows that eve if

More information

4.3 Growth Rates of Solutions to Recurrences

4.3 Growth Rates of Solutions to Recurrences 4.3. GROWTH RATES OF SOLUTIONS TO RECURRENCES 81 4.3 Growth Rates of Solutios to Recurreces 4.3.1 Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer.

More information

2.4.2 A Theorem About Absolutely Convergent Series

2.4.2 A Theorem About Absolutely Convergent Series 0 Versio of August 27, 200 CHAPTER 2. INFINITE SERIES Add these two series: + 3 2 + 5 + 7 4 + 9 + 6 +... = 3 l 2. (2.20) 2 Sice the reciprocal of each iteger occurs exactly oce i the last series, we would

More information

Section 8.5 Alternating Series and Absolute Convergence

Section 8.5 Alternating Series and Absolute Convergence Sectio 85 Alteratig Series ad Absolute Covergece AP E XC II Versio 20 Authors Gregory Hartma, PhD Departmet of Applied Mathema cs Virgiia Military Is tute Bria Heiold, PhD Departmet of Mathema cs ad Computer

More information

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j The -Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discrete-time LTI systems. 7.

More information

Solutions to Homework 7

Solutions to Homework 7 Solutios to Homework 7 Due Wedesday, August 4, 004. Chapter 4.1) 3, 4, 9, 0, 7, 30. Chapter 4.) 4, 9, 10, 11, 1. Chapter 4.1. Solutio to problem 3. The sum has the form a 1 a + a 3 with a k = 1/k. Sice

More information

Math 25 Solutions to practice problems

Math 25 Solutions to practice problems Math 5: Advaced Calculus UC Davis, Sprig 0 Math 5 Solutios to practice problems Questio For = 0,,, 3,... ad 0 k defie umbers C k C k =! k!( k)! (for k = 0 ad k = we defie C 0 = C = ). by = ( )... ( k +

More information

CALCULUS II. Sequences and Series. Paul Dawkins

CALCULUS II. Sequences and Series. Paul Dawkins CALCULUS II Sequeces ad Series Paul Dawkis Table of Cotets Preface... ii Sequeces ad Series... 3 Itroductio... 3 Sequeces... 5 More o Sequeces...5 Series The Basics... Series Covergece/Divergece...7 Series

More information

Convergence of random variables. (telegram style notes) P.J.C. Spreij

Convergence of random variables. (telegram style notes) P.J.C. Spreij Covergece of radom variables (telegram style otes).j.c. Spreij this versio: September 6, 2005 Itroductio As we kow, radom variables are by defiitio measurable fuctios o some uderlyig measurable space

More information

The natural exponential function

The natural exponential function The atural expoetial fuctio Attila Máté Brookly College of the City Uiversity of New York December, 205 Cotets The atural expoetial fuctio for real x. Beroulli s iequality.....................................2

More information

Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests

Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests Sectio.6 Absolute ad Coditioal Covergece, Root ad Ratio Tests I this chapter we have see several examples of covergece tests that oly apply to series whose terms are oegative. I this sectio, we will lear

More information

M17 MAT25-21 HOMEWORK 5 SOLUTIONS

M17 MAT25-21 HOMEWORK 5 SOLUTIONS M17 MAT5-1 HOMEWORK 5 SOLUTIONS 1. To Had I Cauchy Codesatio Test. Exercise 1: Applicatio of the Cauchy Codesatio Test Use the Cauchy Codesatio Test to prove that 1 diverges. Solutio 1. Give the series

More information

A detailed proof of the irrationality of π

A detailed proof of the irrationality of π Matthew Straugh Math 4 Midterm A detailed proof of the irratioality of The proof is due to Iva Nive (1947) ad essetial to the proof are Lemmas ad 3 due to Charles Hermite (18 s) First let us itroduce some

More information

MATH 21 SECTION NOTES

MATH 21 SECTION NOTES MATH SECTION NOTES EVAN WARNER. March 9.. Admiistrative miscellay. These weekly sectios will be for some review ad may example problems, i geeral. Attedace will be take as per class policy. We will be

More information

Chapter 8. Uniform Convergence and Differentiation.

Chapter 8. Uniform Convergence and Differentiation. Chapter 8 Uiform Covergece ad Differetiatio This chapter cotiues the study of the cosequece of uiform covergece of a series of fuctio I Chapter 7 we have observed that the uiform limit of a sequece of

More information