You have probably learned about Taylor polynomials and, in particular, that + +

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1 Sequeces ad Series You have probably leared about Taylor polyomials ad, i particular, that e x = +x+ x2 2! + x x + +!! +E (x) where E (x) is the error itroduced whe you approximate e x by its Taylor polyomial of degree. You may have eve see a formula for E (x). We are ow goig to ask what happes as goes to ifiity? Does the error go zero, givig a exact formula for e x? We shall later see that it does ad that e x x =! But we shall also see other fuctios for which the correspodig error obeys E (x) = 0 for some values of x ad ot for other values of x. Before we ca deal with such questios, we have to build some foudatios. Sequeces Defiitio. A sequece is a list of ifiitely may umbers with a specified order. It is deoted { a, a 2, a,, a, } or { a } or { a } Example 2 Here are three sequeces. {, 2,,, }, { }, 2,,,, { },,,,, ( ), or or or { a = } { } a = { a = ( ) } It is ot ecessary that there be a simple explicit formula for the th term of a sequece. For example the decimal digits of π is a perfectly good sequece. {,, 4,, 5, 9, 2, 6, 5,, 5, 8, 9, 7, 9,, 2,, 8, 4, 6, 2, 6, 4,,, 8, } Example 2 Our primary cocer with sequeces will be the behaviour of a as teds to ifiity ad, i particular, whether or ot a settles dow to some value as teds to ifiity. Sequeces ad Series March 8, 206

2 Defiitio. A sequece { a } is said to coverge to the it A if a approaches A as teds to ifiity. If so, we write a = A or a A as A sequece is said to coverge if it coverges to some it. Otherwise it is said to diverge. Example 4 Three of the four sequeces i Example 2 diverge: The sequece { a = } diverges because a grows without boud, rather tha approachig some fiite value, as teds to ifiity. The sequece { a = ( ) } diverges because a oscillates betwee + ad rather tha approachig a sige value as teds to ifiity. The sequece of the decimal digits of π also diverges, though the proof that this is the case is a bit beyod us right ow. The other sequece i Example 2 has a =. As teds to ifiity, = 0 teds to zero. So Example 4 Example 5 ( ) 2+ Here is a little less trivial example. To study the behaviour of as, it is a good 2+ idea to write it as 2+ = 2+ As, the i the deomiator teds to zero, so that the deomiator 2+ teds to 2 ad teds to. So = 2+ = 2 Example 5 You have already had a fair bit of experiece dealig with its like x f(x). This experiece ca be easily trasferred to dealig with a its by usig the followig result. Sequeces ad Series 2 March 8, 206

3 Theorem 6. If f(x) = L x ad if a = f() for all positive itegers, the a = L Example 7 ( e ) Set f(x) = e x. The e = f() ad x e x = 0 = e = 0 Example 7 The bulk of the rules that you have used to work with its like x f(x) also apply to its like a. Theorem 8 (Arithmetic of its). Let A, B ad C be real umbers ad let the two sequeces { a } ad { b } coverge to A ad B respectively. That is, assume that a = A b = B The the followig its hold. (a) [ a +b ] = A+B (The it of the sum is the sum of the its.) (b) [ a b ] = A B (The it of the differece is the differece of the its.) (c) Ca = CA. (d) a b = AB (The it of the product is the product of the its.) a (e) If B 0 the = A b B (The it of the quotiet is the quotiet of the its provided the it of the deomiator is ot zero.) Sequeces ad Series March 8, 206

4 Example 9 Combiig Examples 5 ad 7, [ ] 2+ +7e = = e by Theorem 8.a e by Theorem 8.c = +7 0 by Examples 5 ad 7 2 = 2 Example 9 Theorem 0 (Squeeze theorem). If a c b for all atural umbers, ad if a = b = L the c = L Example I this example we use the squeeze theorem to evaluate [ + π ] where π is the th decimal digit of π. Precisely π = π 2 = π = 4 π 4 = π 5 = 5 π 6 = 9 We do ot have a simple formula for π. But we do kow that 0 π 9 = 0 π 9 = + π + 9 ad we also kow that [ = + 9 ] = So the squeeze theorem with a =, b = + π, ad c = + 9 gives [ + π ] = Example Sequeces ad Series 4 March 8, 206

5 Theorem 2 (Cotiuous fuctios of its). If a = L ad if the fuctio g(x) is cotiuous at L, the g(a ) = g(l) Example ( si π 2+ ) Write si π 2+ = g( 2+) with g(x) = si(πx). We saw, i Example 5 that 2+ = 2 Sice g(x) = si(πx) is cotiuous at x =, which is the it of, we have 2 2+ π ( ) ( si 2+ = g = g = si 2+ 2) π 2 = Example Series A series is a sum a +a 2 +a + +a + of ifiitely may terms. I summatio otatio, it is writte A example is the decimal expasio of, which you will recall is 0.. You will also recall that 0. meas a = 0 The summatio idex is ofcourse a dummy idex. Youcause ay symbol you like (withi reaso) for the summatio idex. 0 = i= 0 = i j= 0 = j l= 0 l Sequeces ad Series 5 March 8, 206

6 A series ca be expressed usig summatio otatio i may differet ways. For example j=2 l=0 0 + =2 0 = 0 j = 0 l+ = {}}{ 0 + j=2 {}}{ 0 + l=0 {}}{ = 0 + =2 {}}{ 00 + j= {}}{ 00 + l= {}}{ 00 + =2 {}}{ 00 + = {}}{ j=4 {}}{ l= {}}{ = {}}{ all represet exactly the same series. To get from the first lie to the secod lie, substitute = j everywhere, icludig i the its of summatio (so that = becomes j = which is rewritte as j = 2). To get from the first lie to the third lie, substitute = l+ everywhere, icludig i the its of summatio (so that = becomes l + = which is rewritte as l = 0). Wheever you are i doubt as to what series a summatio otatio expressio represets, write out the first few terms, as above. Of course a sum of ifiitely may terms may or may ot add up to a fiite umber. Here are preiary examples of each of those two possibilities. Example 4 ( ) 0 As we have just see above the series 0 = Notice that the th term i that sum is {}}{ 0 + is 0 =2 {}}{ 00 + = {}}{ zeroes {}}{ 0 = So the sum of the first 5, 0, 5 ad 20 terms i that series are = = = 0. 0 = 0. It sure looks like that, as we add more ad more terms, we get closer ad closer to 0. =. So it is very reasoable to defie to be. 0 Sequeces ad Series 6 March 8, 206

7 Example 4 Example 5 ( ad ( ) ) Every term i the series is exactly. So the sum of the first N terms is exactly N. As we add more ad more terms this grows uboudedly. So it is very reasoable to say that the series diverges. The series = =5 =2 =4 {}}{{}}{{}}{{}}{{}}{ ( ) = ( )+ + ( )+ + ( )+ So the sum of the first N terms is 0 if N is eve ad if N is odd. As we add more ad more terms from the series, the sum alterates betwee 0 ad for ever ad ever. So the sum of all ifiitely may terms does ot make ay sese ad it is agai reasoable to say that the series ( ) diverges. Example 5 I geeral, to defie what we mea by the sum a of ifiitely may terms, we approximate it by a fiite sum, say of N terms, ad take the it as N teds to ifiity. Here are the associated formal defiitios. Defiitio 6. The N th partial sum of the series a is S N = If the sequece { } S N coverges as N, say to S, the we say that the series N= a coverges ad we write N a a = S If the sequece of partial sums diverges, we say that the series diverges. Example 7 (Geometric Series) Let a ad r be ay two fixed real umbers with a 0. The series a+ar +ar 2 + +ar + = ar is called the geometric series with first term a ad ratio r. Note that we have chose to start the summatio idex at = 0. That s fie. The first term is the = 0 term, which Sequeces ad Series 7 March 8, 206

8 is ar 0 = a. The secod term is the = term, which is ar = ar. Ad so o. We could have also writte the series ar. That s exactly the same series the first term is ar = ar = a, the secod term is ar =2 = ar 2 = ar, ad so o. Regardless of how a geometric series is writte, a is the first term ad r is the ratio betwee successive terms. The partial sums of ay geometric series ca be computed exactly. Defie S N = N ar = a+ar +ar 2 + +ar N The secret to evaluatig this sum is to see what happes whe we multiply it r: rs N = r ( a+ar +ar 2 + +ar N) = ar +ar 2 +ar + +ar N+ This is almost the same as S N. The oly differeces are that the first term, a, is missig ad oe additioal term, ar N+, has bee tacked o the ed. So rs N = S N a+ar N+ ( r)s N = a ( r N+) If r, we ca ow solve for S N just by dividig the r across. If r =, S N is exactly N + copies of a added together. So a rn+ if r r S N = a(n +) if r = If r <, the r N+ teds to zero as N, so that S N coverges to r as N ad ar = a r if r < O the other had if r, S N diverges because if r >, the r N grows to as N. If r <, the the magitude of r N grows to, ad the sig of r N oscillates betwee + ad, as N. If r = +, the N + grows to as N. If r =, the r N just oscillates betwee + ad as N. So if r the geometric series ar diverges. Example 7 Sequeces ad Series 8 March 8, 206

9 Example 8 (Decimal Expasios) The decimal expasio 0. = = 0 is a geometric series with the first term a = ad the ratio r =. So, by Example 7, = 0 = /0 /0 = /0 9/0 = just as we would have expected. Similarly, = is a geometric series with the first term a = = ad the ratio r =. So, by Example 7, = 6/00 /00 = 6 /00 99/00 agai, as expected. I this way ay periodic decimal expasio coverges to a ratio of two itegers that is, to a ratioal umber. Here is aother more complicated example. = = = =2 = = = / by Example 7 with a = ad r = 00 Example 8 Example 9 (Telescopig Series) I this example we are goig to study the series. This is a rather artificial series (+) that has bee rigged to illustrate a pheomeo call telescopig. Because (+) = + Sequeces ad Series 9 March 8, 206

10 we ca compute the partial sums for this series exactly. S N = N (N +) ( = 2) ( + 2 ( + ) + + 4) ( N ) N + The secod term of each bracket exactly cacels the first term of the followig bracket. So the sum telescopes leavig just S N = N + ad we ca ow easily compute ( (+) = S N = ) = N N N + Example 9 The usual additio ad multiplicatio by costats rules also apply to series. Theorem 20 (Arithmetic of series). Let A, B ad C be real umbers ad let the two series a ad b coverge to A ad B respectively. That is, assume that a = A b = B The the followig hold. (a) (b) [ ] [ ] a +b = A+B ad a b = A B Ca = CA. Example 2 As a simple example of how we use the arithmetic of series Theorem 20, cosider [ ] (+) We recogize that we kow how to compute parts of this sum. We kow that 7 = /7 = /7 6 Sequeces ad Series 0 March 8, 206

11 because it is a geometric series (Example 7) with first term a = ad ratio r =. Ad we 7 7 kow that (+) = by Example 9. We ca ow use Theorem 20 to build the specified complicated series out of these two simple pieces. [ ] = (+) = (+) 7 +2 (+) = 6 +2 = 6 by Theorem 20.a by Theorem 20.b Example 2 Covergece Tests It is very commo to ecouter series for which it is difficult, or eve virtually impossible, to determie the sum exactly. Ofte you try to evaluate the sum approximately by trucatig it, i.e. havig the idex ru oly up to some fiite N, rather tha ifiity. But there is o poit i doig so if the series diverges. So you like to at least kow if the series coverges or diverges. Furthermore you would also like to kow what error is itroduced whe you approximate a by the trucated series N a. That s called the trucatio error. There are a umber of covergece tests to help you with this. The Divergece Test Our first test is very easy to apply, but it is also rarely useful. It just allows us to quickly reject some trivially diverget series. It is based o the observatio that by defiitio, a series a coverges to S whe the partial sums S N = N a coverge to S. The, as N, we have S N S ad, because N too, we also have S N S. So a N = S N S N S S = 0. Theorem 22 (Divergece Test). If the sequece { a } fails to coverge to zero as, the the series a diverges. Sequeces ad Series March 8, 206

12 Example 2 Let a = +. The a = + = + / = 0 So the series diverges. + Example 2 Warig 24 Thedivergecetestisa oewaytest. Ittellsusthatif a isozero,orfailstoexist, the the series a diverges. But it tells us absolutely othig whe a = 0. I particular, it is perfectly possible for a series a to diverge eve though a = 0. A example is. We ll show i Example 26, below, that it divergesẇarig 24 The Itegral Test Theorem 25 (The Itegral Test). Let N 0 be ay atural umber. If f(x) is a fuctio which is defied ad cotiuous for all x N 0 ad which obeys (i) f(x) 0 for all x N 0 ad (ii) f(x) decreases as x icreases ad (iii) f() = a for all N 0. y = f(x) y a a 2 a a x The a coverges N 0 f(x) dx coverges Furthermore, whe the series coverges, the trucatio error a N a N f(x) dx for all N N 0 Sequeces ad Series 2 March 8, 206

13 Proof. Let I be ay fixed iteger with I > N 0. The a coverges if ad oly if =I a coverges removig a fixed fiite umber of terms from a series caot impact whether or ot it coverges. Sice a 0 for all I > N 0, the sequece of partial sums s l = l =I a obeys s l+ = s l +a + s l. That is, s l icreases as l icreases. So { s l } must either coverge to some fiite umber or icrease to ifiity. That is, either =I a coverges to a fiite umber or it is +. y = f(x) a I I a I+ I + a I+2 I +2 a I+ I + x Look at the figure above. The shaded area i the figure is =I a because the first shaded rectagle has height a I ad width, ad hece area a I ad the secod shaded rectagle has height a I+ ad width, ad hece area a I+, ad so o This shaded area is smaller tha the area uder the curve y = f(x) for I x <. So a =I I ad, if the itegral is fiite, the sum =I a is fiite too. y = f(x) f(x) dx () a I I a I+ I + a I+2 I +2 a I+ I + x For the divergece case look at the figure above. The (ew) shaded area i the figure is agai =I a because the first shaded rectagle has height a I ad width, ad hece area a I ad the secod shaded rectagle has height a I+ ad width, ad hece area a I+, ad so o Sequeces ad Series March 8, 206

14 This time the shaded area is larger tha the area uder the curve y = f(x) for I x <. So a f(x) dx =I ad, if the itegral is ifiite, the sum =I a is ifiite too. Fially, the boud o the tructio error is just the special case of () with I = N +: a N a = I =N+ a N f(x) dx Example 26 ( ) p Let p > 0. We ll ow use the itegral test to determie whether or ot the series (which is sometimes called the p series) coverges. To do so, we eed a fuctio f(x) that obeys f() = a = for all bigger tha p some N 0. Certaily f(x) = obeys f() = for all. So let s pick this f ad x p p try N 0 =. (We ca always icrease N 0 later if we eed to.) This fuctio also obeys the other two coditios of Theorem 25: (i) f(x) > 0 for all x N 0 = ad (ii) f(x) decreases as x icreases because f (x) = p x p+ < 0 for all x N 0 =. So the itegral test tells us that the series dx coverges. x p p coverges if ad oly if the itegral p We have already see, i Example 4 of the otes Improper Itegrals, that the itegral dx coverges if ad oly if p >. x p So we coclude that coverges if ad oly if p >. p I particular, the series, which is called the harmoic series, has p = ad so diverges. As we add more ad more terms of this series together, the terms we add, amely, get smaller ad smaller ad ted to zero, but the rate at which they ted to zero is small eough that the full sum is still ifiite. O the other had, the series has p = > ad so coverges This time as we add more ad more terms of this series together, the terms we add, amely, ted to zero (just) fast eough that the full sum is fiite. Mid you, for this example, the covergece takes place very slowly you have to take a huge umber of terms to get a decet approximatio to the full sum. If we approximate by the trucated series N.00000, we make a error of at most dx R dx = = x R x [ R R ] = N N N N Sequeces ad Series 4 March 8, 206

15 This does ted to zero as N, but really slowly. Example 26 Example 27 ( =2 ) (log) p Let p > 0. We ll ow use the itegral test to determie whether or ot the series coverges. =2 (log) p As i the last example, we start by choosig a fuctio that obeys f() = a = (log) p for all bigger tha some N 0. Certaily f(x) = obeys f() = for all x(logx) p (log) p 2. So let s use that f ad try N 0 = 2. Now let s check the other two coditios of Theorem 25: (i) Both x ad logx are positive for all x >, so f(x) > 0 for all x N 0 = 2. (ii) As x icreases both x ad logx icrease ad so x(logx) p icreases ad f(x) decreases. Sotheitegraltest tellsusthattheseries 2 dx x(logx) p coverges. =2 coverges ifadolyiftheitegral (log) p To test the covergece of the itegral, we make the substitutio u = logx, du = dx x. R 2 dx logr x(logx) = du p log2 u p Wealreadykowthattheitegral theitegral coverges if ad oly if p >. So we coclude that coverges if ad oly if p >. (log) p du, adhecetheitegral R u p 2 Example 27 dx, x(logx) p The Compariso Test Our ext covergece test is the compariso test. It is much like the compariso test for improper itegrals (see Theorem 2 of the otes Improper Itegrals ) ad is true for much the same reasos. Sequeces ad Series 5 March 8, 206

16 Theorem 28 (The Compariso Test). Let N 0 be a atural umber ad let K > 0. (a) If a Kc for all N 0 ad c coverges, the a coverges. (b) If a Kd 0 for all N 0 ad d diverges, the a diverges. Proof. We will ot prove this theorem. We ll just observe that it is very reasoable. That s why there are quotatio marks aroud Proof. (a) If c coverges to a fiite umber ad if the terms i a are smaller tha the terms (b) If i c, the it is o surprise that a coverges too. d diverges (i.e. adds up to ) ad if the terms i a are larger tha the terms i d, the of course a adds up to, ad so diverges, too. The compariso test for series is also used i much the same way as is the compariso test for improper itegrals. Example 29 ( ) We could determie whether or ot the series coverges by applyig the itegral test. But it is ot worth the effort. Whether or ot ay series coverges is determied by the behaviour of the summad for very large. So the first step i tacklig such a problem is to develop some ituitio about the behaviour of a whe is very large. Step : Develop ituitio. I this case, whe is very large 2 2 so that. We already kow, from Example 26, that coverges if ad p oly if p >. So, which has p = 2, coverges, ad we would expect that 2 coverges too Step 2: Verify ituitio. We ca use the compariso test to cofirm that this is ideed the case. For ay, > 2, so that. So the compariso test, Theorem 28, with a = ad c =, tells us that 2 coverges. 2 +2x+ Example 29 The symbol meas much larger tha. Sequeces ad Series 6 March 8, 206

17 Of course the previous example was rigged to give a easy applicatio of the compariso test. It is ofte relatively easy, usig argumets like those i Example 29, to fid a simple series b with b almost the same as a whe is large. However it is pretty rare that a b for all. It is much more commo that a Kb for some costat K. This is eough to allow applicatio of the compariso test. Here is a example. Example 0 ( +cos ) / As i the previous example, the first step is to develop some ituitio about the behaviour of a whe is very large. Step : Develop ituitio. Whe is very large, cos so that the umerator +cos ad / so that the deomiator /. So whe is very large a = +cos / = 2 We already kow from Example 26, with p = 2, that expect that coverges too. +cos / coverges, so we would 2 Step 2: Verify ituitio. We ca use the compariso test to cofirm that this is ideed the case. To do so we eed to fid a costat K such that a = +cos = +cos is / / smaller tha K for all. A good way to do that is to factor the domiat term (i 2 this case ) out of the umerator ad also factor the domiat term (i this case ) out of the deomiator. a = +cos / = + cos = cos + 2 So ow we eed to fid a costat K such that + (cos)/ / is smaller tha K for all. First cosider the umerator +(cos). For all ad cos So the umerator +(cos) is always smaller tha +() = 2. Next cosider the deomiator /. Whe =, / = 2/ = 2. As icreases, icreases, so / decreases towards 0, so / icreases towards ad / decreases towards. Cosequetly / is ever bigger tha 2 Sequeces ad Series 7 March 8, 206

18 Asthe umerator +(cos) isalways smaller tha2ad isalways smaller / tha, the fractio 2 + cos ) 2( = 2 We ow kow that a = / / 2 ad the compariso test tells us that +cos coverges. / Example 0 The last example was actually a relatively simple applicatio of the compariso theorem fidig a suitable costat K ca be really tedious. Fortuately, there is a variat of the compariso test that completely eiates the eed to fid K. Theorem (Limitig Compariso Theorem). Let a ad b be two series with b > 0 for all. Assume that exists. a = L b (a) If b coverges, the a coverges too. (b) If L 0 ad b diverges, the a diverges too. I particular, if L 0, the a coverges if ad oly if b coverges. Proof. We will ot prove this theorem, but we will explai the idea behid the proof. a (a) Because we are told that b = L, we kow that, whe is large, a b is very close to L, so that a b is very close to L. I particular, there is some atural umber N so that a b L +, ad hece a Kb with K = L +, for all N. The compariso Theorem 28 ow implies that a coverges. (b) Let s suppose that L > 0. (If L < 0, just replace a with a.) Because we are told that a b = L, we kow that, whe is large, a b is very close to L. Sequeces ad Series 8 March 8, 206

19 I particular, there is some atural umber N so that a b L, ad hece 2 a Kb with K = L 2 > 0, for all N. The compariso Theorem 28 ow implies that a diverges. Example 2 ( ) Set a = We first try to develop some ituitio about the behaviour of a for large ad the we cofirm that our ituitio was correct. Step : Develop ituitio. Whe, the umerator +, ad the deomiator so that a = ad it looks like our series should 2 /2 coverge by Example 26 with p =. 2 Step 2: Verify ituitio. To cofirm our ituitio we set b = ad compute the /2 it a = b /2 / = Agai it is a good idea to factor the domiat term out of the umerator ad the domiat term out of the deomiator. a 2 + / + / = b 2( ) = 2 /+ / 2 2 /+ = / 2 We already kow that the series b = coverges by Example 26 with /2 p =. So our series coverges by the itig compariso test, Theorem. 2 Example 2 The Alteratig Series Test Whe the sigs of successive terms i a series alterate betwee + ad, like for example i + +, the series is called a alteratig series. More geerally, the series 2 4 a a 2 +a a 4 + = ( ) a is alteratig if every a 0. Ofte (but ot always) the terms i alteratig series get successively smaller. That is, the a a 2 a. I this case: The first partial sum is S = a. The secod partial sum, S 2 = a a 2, is smaller tha S by a 2. Sequeces ad Series 9 March 8, 206

20 The third partial sum, S = S 2 +a, is bigger tha S 2 by a, but because a a 2, S remais smaller tha S. See the figure below. The fourth partial sum, S 4 = S a 4, is smaller tha S by a 4, but because a 4 a, S 4 remais bigger tha S 2. Agai, see the figure below. Ad so o. So the successive partial sums oscillate, but with ever decreasig amplitude. If, i additio, a teds to 0 as teds to, the amplitude of oscillatio teds to zero ad the sequece S, S 2, S, coverges. This is illustrated i the figure S = a a 2 +a a 4 S 2 S S 4 S 5 S 6 S 7 S N Here is a covergece test for alteratig series that exploits this structure, ad that is really easy to apply. Theorem (Alteratig Series Test). Let { a } be a sequece of real umbers that obeys (i) a 0 for all ad (ii) a + a for all (i.e. the sequece is mootoe decreasig) ad (iii) a = 0. The a a 2 +a a 4 + = ( ) a = S coverges ad, for each atural umber N, S S N is betwee 0 ad (the first dropped term) ( ) N a N+. Proof. We are ot goig to give a complete proof. We shall fix ay atural umber N ad cocetrate o the last statemet, which gives a boud o the trucatio error (which is the Sequeces ad Series 20 March 8, 206

21 error itroduced whe you approximate the full series by the partial sum S N ) E N = S S N = =N+ ] ( ) a = ( ) [a N N+ a N+2 +a N+ a N+4 + This is of course aother series. We re goig to study the partial sums for that series. If l > N +, with l N eve, S N,l = l =N+ ( ) a {}}{{}}{{}}{ ( ) N S N,l = (a N+ a N+2 )+(a N+ a N+4 )+ +(a l a l ) 0 ad 0 0 {}}{ ( ) N {}}{ S N,l+ = S N,l + a l+ 0 This tells us that ( ) N S N,l 0 for all l > N +. Similarly, if l > N +, with l N odd, {}}{{}}{{}}{ ( ) N S N,l = a N+ ( a N+2 a N+ ) ( a N+4 a N+5 ) (a l a l ) a N+ ad a N+ {}}{ 0 ( ) N {}}{ S N,l+ = S N,l a l+ a N+ This tells us that ( ) N S N,l a N+ for all for all l > N +. So we ow kow that S N,l lies betwee its first term, ( ) N a N+, ad 0 for all l > N +. While we are ot goig to prove it, this implies that, sice a N+ 0 as N, the series S coverges ad that S S N = l S N,l lies betwee ( ) N a N+, ad 0. Example 4 We have already see, i Example 26, that the harmoic series diverges. O the other had, the series ( ) coverges by the alteratig series test with a =. Note that (i) a = 0 for all, so that ( ) (ii) a = decreases as icreases, ad (iii) a = = 0. really is a alteratig series, ad Sequeces ad Series 2 March 8, 206

22 so that all of the hypotheses of the alteratig series test, i.e. of Theorem, are satisfied. Example 4 Example 5 (e) You may already kow that e x = x. I ay evet, we shall prove this i Example 6,! below. I particular e = e = ( ) =!! + 2!! + 4! 5! + is a alteratig series ad satisfies all of the coditios of the alteratig series test, Theorem a: (i) The terms i the series alterate i sig. (ii) The magitude of the th term i the series decreases mootoically as icreases. (iii) The th term i the series coverges to zero as. So the alteratig series test guaratees that, if we approximate, for example, e 2!! + 4! 5! + 6! 7! + 8! 9! the the error i this approximatio lies betwee 0 ad the ext term i the series, which is 0!. That is 2!! + 4! 5! + 6! 7! + 8! 9! e 2!! + 4! 5! + 6! 7! + 8! 9! + 0! so that e !! 4! 5! 6! 7! 8! 9! 0! 2!! 4! 5! 6! 7! 8! 9! which, to seve decimal places says (To seve decimal places e = ) e Example 5 The Ratio Test Theorem 6 (Ratio Test). Let N be ay positive iteger ad assume that a 0 for all N. a (a) If + a = L <, the a coverges. a (b) If + a = L >, or a + a = +, the a diverges. Sequeces ad Series 22 March 8, 206

23 Proof. (a) Pick ay umber R obeyig L < R <. We are assumig that a + a approaches L as. I particular there must be some atural umber M so that a + a R for all M. So a + R a for all M. I particular a M+ R a M a M+2 R a M+ R 2 a M a M+ R a M+2 R a M. a M+l R l a M for all l 0. The series l=0 Rl a M is a geometric series with ratio R smaller tha oe i magitude ad so coverges. Cosequetly, by the compariso test with a replaced by A l = a +l ad c replaced by C l = R l a M, the series a M+l = a coverges. So the l= =M+ series a coverges too. (b) We are assumig that a + a approaches L > as. I particular there must be some atural umber M > N so that a + a for all M. So a+ a for all M. That is, a icreases as icreases as log as M. So a a M for all M ad a caot coverge to zero as. So the series diverges by the divergece test. Warig 7 Beware that the ratio test provides absolutely o coclusio about the covergece or divergece of the series a a if + a =. See Example 40, below. Warig 7 Example 8 ( ax ) Fix ay two ozero real umbers a ad x. We have already see i Example 7 we have just reamed r to x that the geometric series ax coverges whe x < ad diverges whe x. We are ow goig to cosider a ew series, costructed by differetiatig 2 each term i the geometric series ax. This ew series is with a = ax a Let s apply the ratio test. a + a(+)x + = = (+ a ax x = ) x L = x as 2 We shall see later, i Theorem 56, that the fuctio ax is ideed the derivative of the fuctio ax. Sequeces ad Series 2 March 8, 206

24 The ratio test ow tells us that the series ax coverges if x < ad diverges if x >. It says othig about the cases x = ±. But i both of those cases a = a(±) does ot coverge to zero as ad the series diverges by the divergece test. Example 8 Example 9 ( a + X+ ) Oce agai, fix ay two ozero real umbers a ad X. We agai start with the geometric series ax but this time we costruct a ew series by itegratig each term, ax, from x = 0 to x = X givig a + X+. The resultig ew series is a with a = a + X+ To apply the ratio test we eed to compute a a + = +2 X+2 a a = + +2 X = + X L = X as + X+ + 2 The ratio test ow tells us that the series a + X+ coverges if X < ad diverges if X >. It says othig about the cases X = ±. If X =, the series reduces to a + X+ = X= a + = a m m= with m = + which is just a times the harmoic series, which we kow diverges, by Example 26. If X =, the series reduces to a + X+ = ( ) + a X= + which coverges by the alteratig series test. See Example 4. I coclusio, the series a + X+ coverges if ad oly if X <. Example 9 Example 40 (L = ) I this example, we are goig to see three differet series that have a + a =. Oe is goig to diverge ad the other two are goig to coverge. The first series is the harmoic series a with a = We shall also see later, i Theorem 56, that the fuctio the fuctio ax. a + x+ is ideed a atiderivative of Sequeces ad Series 24 March 8, 206

25 We have already see, i Example 26, that this series diverges. It has a + = a = + = L = as + + The secod series is the alteratig harmoic series a with a = ( ) We have already see, i Example 4, that this series coverges. But it also has a + ( ) = + a = + = L = as ( ) + The third series is a with a = 2 We have already see, i Example 26 with p = 2, that this series coverges. But it also has a + = (+) 2 a = 2 (+) = 2 (+ L = as 2 )2 Example 40 Covergece Test List We ow have half a doze covergece tests: Divergece Test works well whe the th term i the series fails to coverge to zero as teds to ifiity Alteratig Series Test works well whe successive terms i the series alterate i sig do t forget to check that successive terms decrease i magitude ad ted to zero as teds to ifiity Itegral Test works well whe substitutig x for i the th term gives a fuctio, f(x), that you ca itegrate do t forget to check that f(x) 0 ad that f(x) decreases as x icreases Sequeces ad Series 25 March 8, 206

26 Ratio Test workswellwhe a + a simplifieseoughthatyoucaeasilycompute a + a = L this ofte happes whe a cotais powers, like 7, or factorials, like! do t forget that L = tells you othig about the covergece/divergece of the series Compariso Test ad Limitig Compariso Test works well whe, for very large, the th term a is approximately the same as a simpler term b (see Example 0) do t forget to check that b 0 usually the Limitig Compariso Test is easier to apply tha the Compariso Test Absolute ad Coditioal Covergece Defiitio 4 (Absolute ad coditioal covergece). (a) A series a is said to coverge absolutely if the series a coverges. (b) If a coverges but a diverges we say that a is coditioally coverget. Theorem 42 (Absolute covergece implies covergece). If theseries a coverges thetheseries a alsocoverges. That is, absolute covergece implies covergece. Recall that some of our covergece tests (for example, the itegral test) may oly be applied to series with positive terms. Theorem 42 opes up the possibility of applyig positive oly covergece tests to series whose terms are ot all positive, by checkig for absolute covergece rather tha for plai covergece. Example 4 ( ( ) ) The alteratig harmoic series ( ) series test). But the harmoic series the alteratig harmoic series ( ) of Example 4 coverges (by the alteratig of Example 26 diverges (by the itegral test). So coverges coditioally. Sequeces ad Series 26 March 8, 206

27 Example 4 Example 44 ( ( ) ) 2 Because the series ( ) = of Example 26 coverges (by the itegral test), 2 2 the series ( ) coverges absolutely, ad hece coverges. 2 Example 44 Example 45 (radom sigs) Imagie flippig a coi ifiitely may times. Set σ = + if the th flip comes up heads ad σ = if the th flip comes up tails. The series is ot i geeral a ( )σ 2 alteratig series. But we kow that the series ( ) σ = coverges. So 2 2 coverges absolutely, ad hece coverges. ( )σ 2 Example 45 The Delicacy of Coditioally Coverget Series (Optioal) Coditioally coverget series have to be treated with great care. For example, switchig the order of the terms i a fiite sum does ot chage its value = The same is true for absolutely coverget series. But it is ot true for coditioally coverget series. I fact by reorderig ay coditioally coverget series, you ca make it add up to ay umber you like, icludig + ad. Here is a example that shows why. Example 46 I this example, we ll reorder the coditioally coverget series ( ) so that it adds up to exactly.24. (Of course, the target.24 has bee chose at radom. It ca be replaced by ay umber you like.) First create two lists of umbers the first list cosistig of the positive terms of the series, i order, ad the secod cosistig of the egative umbers of the series, i order.,, 5, 7, 2, 4, 6, Note that if we add together the umbers i the secod list, we get 2[ ], 2 which is just times the harmoic series. So the umbers i the secod list add up to. 2 Also, if we add together the umbers i the first list, we get which is bigger 5 7 tha = [ ]. So the umbers i the first list add up to Sequeces ad Series 27 March 8, 206

28 Now we build up our reordered series. Start by movig just eough umbers from the begiig of the first list ito the reordered series to get a sum bigger tha =. Next move just eough umbers from the begiig of the secod list ito the reordered series to get a umber less tha = 0.8 Next move just eough umbers from the begiig of the remaiig part of the first list ito the reordered series to get a umber bigger tha =.287 Next move just eough umbers from the begiig of the remaiig part of the secod list ito the reordered series to get a umber less tha =.07 Just keep goig like this. At the ed of each step, the differece betwee the sum ad.24 is smaller tha the magitude of the first uused umber i the list. Sice the umbers i both lists ted to zero as you go farther ad farther up the list, this procedure will geerate a series whose sum is exactly.24. Sice i each step we remove at least oe umber from a list ad we alterate betwee the two lists, the reordered series will cotai all of the terms from, with each term appearig exactly oce. ( ) Example 46 Power Series Remember that we set as our goal, i studyig sequeces ad series, the developmet of machiery which would allow us to aswer questios like, Is e x = x?. We are ow! ready to start workig o series like x. We ll start with the defiitio of a power series.! Sequeces ad Series 28 March 8, 206

29 Defiitio 47. A series of the form A 0 +A (x c)+a 2 (x c) 2 +A (x c) + = A (x c) is called a power series i (x c) or a power series cetered about c. The umbers A are called the coefficiets of the power series. Ofte c = 0 ad the the series reduces to A 0 +A x+a 2 x 2 +A x + = A x The x i a power series is to be thought of as a variable. So each power series is really a whole family of series a differet series for each value of x. Oe possible value of x is x = c ad the the series reduces 4 to A (x c) x=c = {}}{ A (c c) = A 0 + =2 {}}{{}}{ = {}}{ 0 + which trivially coverges to A 0. Let s apply the ratio test to try ad determie which for other values of x this series also coverges. The th term i the series A (x c) is a = A (x c). So the ratio test tells us to compute a + a = A + (x c) + A (x c) = A + x c A Now we are to try ad take the it. There are several possibilities. A If the it + A exists ad equals some ozero value, say A, the the ratio test says that the series A (x c) coverges whe A x c <, i.e. whe x c <, ad diverges whe A x c >, i.e. whe x c >. This A A R = A = [ is called the radius of covergece of the series. A ] + (2) A A If the it + A exists ad equals zero, the A + x A c = 0 for every x ad the ratio test tells us that the series A (x c) coverges for every umber x. I this case we say that the series has a ifiite radius of covergece. If the A + A teds to + as 0, the A + x c A = + for every x c ad the ratio test tells us that the series A (x c) diverges for every umber 4 By covetio, whe 0 0 appears i a power series, it is assiged the value. Sequeces ad Series 29 March 8, 206

30 x c. As we have see above, whe x = c, the series reduces to A , which of course coverges. I this case we say that the series has radius of covergece zero. If A + A does ot approach a it as, the we lear othig from the ratio test. All of these possibilities do happe. We give a example of each below. But first, the cocept of radius of covergece is importat eough to warrat a official defiitio. Defiitio 48. (a) Let 0 < R <. If A (x c) coverges for x c < R, ad diverges for x c > R, the we say that the series has radius of covergece R. (b) If A (x c) coverges for every umber x, we say that the series has a ifiite radius of covergece. (c) If A (x c) diverges for every x c, we say that the series has radius of covergece zero. Example 49 We already kow that, if a 0, the geometric series ax coverges whe x < ad diverges whe x. So, i the termiology of Defiitio 48, the geometric series has radius of covergece R =. As a cosistecy check, we ca also compute R usig (2). The series ax has A = a. So as expected. [ R = ] = A A + [ ] = Example 49 Recall that! = 2 is called factorial. By covetio 0! =. Example 50 The series x has A! =. So! ad A + = A = 0 /(+)! /! =! (+)! = x! has radius of covergece. It coverges for every x. 2 2 (+) = + Sequeces ad Series 0 March 8, 206

31 Example 50 Example 5 The series!x has A =!. So A + (+)! = A! 2 4 (+) = = (+) = ad!x has radius of covergece zero. It coverges oly for x = 0, where it takes the value 0! =. Example 5 Example 52 Let A 0 = 4 ad for each atural umber, let A be oe plus the th decimal digit of π. So every A is a iteger betwee ad 0 ad the series A x = 4+2x+5x 2 +2x +6x 4 +0x 5 + Because π is a irratioal umber A + A caot have a it as. (If you do t kow why this is the case, do t worry about it.) So the ratio test tells us othig about the covergece of this series. But we ca still figure out for which x s it coverges. Because every coefficiet A is o bigger (i magitude) tha 0, the th term i our series obeys A x 0 x ad so is smaller tha the th term i the geometric series 0 x. This geometric series coverges if x <. So, by the compariso test, our series coverges for x < too. Sice every A is at least oe, the th term i our series obeys A x x If x, this a = A x caot coverge to zero as, ad our series diverges by the divergece test. I coclusio, our series coverges if ad oly if x <, ad so has radius of covergece. Example 52 Though we wo t prove it, it is true that every power series has a radius of covergece, A whether or ot the it + A exists. Sequeces ad Series March 8, 206

32 Theorem 5. Let A (x c) be a power series. The oe of the followig alteratives must hold. (i) The power coverges for every umber x. I this case we say that the radius of covergece is. (ii) There is a umber 0 < R < such that the series coverges for x c < R ad diverges for x c > R. The R is called the radius of covergece. (iii) The series coverges for x = 0 ad diverges for all x 0. I this case, we say that the radius of covergece is 0. Defiitio 54. The set of x s for which a power series coverges is called the iterval of covergece for the series. Supposethatthepowerseries A (x c) hasradiusofcovergecer. ThefromTheorem 5, we have that if R =, the its iterval of covergece is < x <, which is also deoted (, ), ad if 0 < R <, the its iterval of covergece must be oe of ad c R < x < c+r, which is also deoted (c R, c+r), or c R x < c+r, which is also deoted [c R, c+r), or c R < x c+r, which is also deoted (c R, c+r], or c R x c+r, which is also deoted [c R, c+r], if R = 0, the its iterval of covergece is just the poit x = 0. Note that kowig the radius covergece, R with 0 < R <, tells you othig about whether or ot the series coverges whe x c = R, i.e. whe x = c ±R. The followig example shows that all four possibilities occur. Example 55 Let p be ay real umber ad cosider the series x. This series has A p = A + A = p (+) p = (+ /) p = p. Sice Sequeces ad Series 2 March 8, 206

33 the series has radius of covergece. So it certaily coverges for x < ad diverges for x >. That just leaves x = ±. So Whe x =, the series reduces to. We kow, from Example 26, that this p series coverges if ad oly if p >. Whex =, theseries reduces to ( ). Bythealteratigseries test, Theorem p, this series coverges wheever p > 0 (so that teds to zero as teds to ifiity). p Whe p 0 (so that does ot ted to zero as teds to ifiity), it diverges by the p divergece test, Theorem 22. (a) The power series x (i.e. p = 0) has iterval of covergece < x <. (b) The power series x (i.e. p = ) has iterval of covergece x <. (c) The power series ( ) x (i.e. p = ) has iterval of covergece < x. (d) The power series x (i.e. p = 2) has iterval of covergece x. 2 Example 55 Workig With Power Series Here is a theorem that ca be used help build power series represetatios for complicated fuctios from power series represetatios of simple fuctios. Sequeces ad Series March 8, 206

34 Theorem 56 (Operatios o Power Series). Assume that the fuctios f(x) ad g(x) are give by the power series f(x) = A (x c) g(x) = B (x c) for all x obeyig x c < R. I particular, we are assumig that both power series have radius of covergece at least R. Also let K be a ozero costat. The f(x)+g(x) = Kf(x) = (x c) N f(x) = x = f (x) = [A +B ](x c) KA (x c) A (x c) +N for ay iteger N A k N (x c) k k=n A (x c) where k = +N (x c) + f(t) dt = A c + [ ] (x c) + f(x) dx = A + C with C a arbitrary costat + for all x obeyig x c < R. I particular, the radius of covergece of each of the six power series o the right had sides is at least R. I fact, if R is the radius of covergece of A (x c), the R is also the radius of covergece of all of the above right had sides, with the possible exceptio of [A +B ](x c). We ll ow use this theorem to build power series represetatios for a buch of fuctios out of the oe simple power series represetatio that we kow the geometric series Example 57 ( x 2+x 2 ) x = x for all x < Fid a power series represetatio for x 2+x 2. Solutio. The secret to fidig power series represetatios for a good may fuctios is to maipulatethemitoaformiwhich appearsadusethegeometricseries represetatio y Sequeces ad Series 4 March 8, 206

35 = y y. We have deliberately reamed the variable to y here it does ot have to be x. We ca do that for the give fuctio. x 2+x 2 = x 2 + x2 /2 = x 2 y y= x 2 /2 = x ( x2 ) 2 2 = = x 2 ( ) x2 /2 = x [ ] y 2 = x 2 y= x 2 /2 ( ) 2 x 2 ( ) 2 + x2+ by Theorem 56, twice = x 2 x 4 + x5 8 x7 6 + This is a perfectly good power series. There is othig wrog with the power of x beig 2 +. I fact, you should try to always write power series i forms that are as easy to uderstad as possible. The geometric series that we used i the secod lie coverges for y < x 2 /2 < x 2 < 2 x < 2 So our power series hasradius of covergece 2 aditerval of covergece 2 < x < 2. Example 57 Example 58 ( ( x) 2 ) Fid a power series represetatio for ( x) 2. Solutio. Oce agai the trick is to express ( x) 2 i terms of x. ( x) = d 2 dx x = d x dx = x by Theorem 56 Note that the = 0 term has disappeared because, for = 0 d dx x = d dx = 0 Our power series has radius of covergece ad iterval of covergece < x <, sice the series diverges for x, by the divergece test. Sequeces ad Series 5 March 8, 206

36 Example 58 Example 59 (log(+x)) Fid a power series represetatio for log( + x). Solutio. Recall that d dx log(+x) = +x so that log(+t) is a atiderivative of +t ad log(+x) = = = x 0 dt +t x 0 = x 0 [ ] ( t) dt ( t) dt by Theorem 56 ( ) x+ + = x x2 2 + x x4 4 Theorem 56 guaratees that the radius of covergece is exactly oe (the radius of covergece of the geometric series ( t) ). Whe x = our series reduces to mius, which is theharmoicseries adso diverges. That s osurprise log(+( )) = + log0 =. Whe x =, the series coverges by the alteratig series test. It is possible to prove, though we wo t do so here, that the sum is log2. So the iterval of covergece is < x. Example 59 Example 60 (arcta x) Fid a power series represetatio for arcta x. Solutio. Recall that d dx arctax = +x 2 so that arctat is a atiderivative of +t 2 ad arctax = = x 0 dt +t 2 = x ( ) x [ ( t 2 ) ] dt = x 0 ( ) t 2 dt Theorem 56 guaratees that the radius of covergece is at least oe (the radius of covergece of the geometric series ( t2 ) ). Whe x = ±, the series coverges by the alteratig series test. Whe x > it diverges by the ratio test or the divergece test. So the iterval of covergece is x. It is possible to prove, though oce agai we wo t do so here, that whe x = ±, the series coverges to arcta(±) = ± π. 4 Example 60 Sequeces ad Series 6 March 8, 206

37 Taylor Series Recall that Taylor polyomials provide a hierarchy of approximatios to a give fuctio f(x) ear a give poit a. The crudest approximatio is the costat approximatio f(x) f(a). The comes the liear, or taget lie, approximatio f(x) f(a)+f (a)(x a). The comes the quadratic approximatio f(x) f(a)+f (a)(x a)+ 2 f (a)(x a) 2. Igeeral, thetaylor polyomial ofdegree, forthefuctiof(x)abouttheexpasio poit a, is the polyomial, T (x), determied by the requiremets that f (m) (a) = T (m) (a) for all 0 m. That is, f ad T have the same derivatives at a, up to order. Explicitly, f(x) T (x) = f(a)+f (a)(x a)+ 2 f (a)(x a) 2 + +! f() (a)(x a) = m=0 m! f(m) (a)(x a) These are of course approximatios ofte very good approximatios ear x = a but still just approximatios. Ca we get exact represetatios by takig the it as? That s the questio we ll cosider ow. Fix a real umber a ad suppose that all derivatives of the fuctio f(x) exist. The, for ay atural umber, f(x) = T (x)+e (x) () where T (x) = f(a)+f (a)(x a)+ +! f() (a)(x a) isthetaylor polyomialofdegreeforthefuctiof(x)adexpasiopoita, ade (x) = f(x) T (x) is the error itroduced whe we approximate f(x) by the polyomial T (x). It is true, though we wo t prove it, that (a) E (x) = (+)! f(+) (c)(x a) + (b) for some (usually ukow) c strictly betwee a ad x. If it happes that, for some x, E (x) teds to zero as, the we have the exact formula f(x) = T (x) for f(x). This is usually writte f(x) =! f() (a)(x a) (4) ad is called the Taylor series of f(x) with expasio poit a. (Whe a = 0 it is also called the Maclauri series of f(x).) It is a power series represetatio for f(x). Sequeces ad Series 7 March 8, 206

38 Example 6 (Expoetial Series) This happes with the expoetial fuctio f(x) = e x. We ll first fid f (m) (0) for all itegers m 0. f(x) = e x f (x) = e x f (x) = e x f(0) = e 0 = f (0) = e 0 = f (0) = e 0 = Applyig () with f(x) = e x ad a = 0, ad usig that f (m) (a) = e a = e 0 = for all m, e x = f(x) = +x+ x2 2! + + x! + (+)! ec x + (5) for some c betwee 0 ad x. Now cosider ay fixed real umber x. As c rus from 0 to x, e c rus from e 0 = to e x. I particular, e c is always betwee ad e x ad so is o bigger the larger of the two umbers ad e x, which is tur is smaller tha +e x. Thus the error term e c E (x) = (+)! x+ [e x +] x + (+)! Let s call e (x) = x + (+)!. We claim that as icreases towards ifiity, e (x) decreases (quickly) towards zero. To see this, let s compare e (x) ad e + (x). e + (x) e (x) = x +2 (+2)! x + (+)! = x +2 So, whe is bigger tha, for example 2 x, we have e +(x) e (x) <. That is, icreasig the 2 idex o e (x) by oe decreases the size of e (x) by a factor of at least two. As a result e (x) must ted to zero as. Cosequetly, for all x, E (x) = 0 ad [ e x = +x+ 2 x2 +! x + + ]! x =! x (6) Example 6 Example 62 (Sie ad Cosie Series) The trigoometric fuctios si x ad cos x also have widely used Taylor series expasios about a = 0. To fid them, we first, compute all derivatives at geeral x. f(x) = six f (x) = cosx f (x) = six f () (x) = cosx f (4) (x) = six g(x) = cosx g (x) = six g (x) = cosx g () (x) = six g (4) (x) = cosx (7) The patter starts over agai with the fourth derivative beig the same as the origial fuctio. Now set x = a = 0. f(x) = six f(0) = 0 f (0) = f (0) = 0 f () (0) = f (4) (0) = 0 g(x) = cosx g(0) = g (0) = 0 g (0) = g () (0) = 0 g (4) (0) = Sequeces ad Series 8 March 8, 206 (8)

39 For six, all eve umbered derivatives (at x = 0) are zero. The odd umbered derivatives alterate betwee ad. For cosx, all odd umbered derivatives are zero. The eve umbered derivatives alterate betwee ad. So, the Taylor polyomials that best approximate six ad cosx ear x = a = 0 are six x! x + 5! x5 cosx 2! x2 + 4! x4 Reviewig (7) we see that every derivative of six ad cosx is oe of ±six ad ±cosx. Cosequetly, whe we apply (b) we always have f (+) (c), g (+) (c) ad hece E (x) x + (+)!. We have already see i Example 6, that x + (+)! (which we called e (x) i Example 6) coverges to zero as. Cosequetly, for both f(x) = six ad f(x) = cosx, we have E (x) = 0 ad Reviewig (8), we coclude that, for all x, [ ] f(x) = f(0)+f (0)x+ +! f() (0)x six = x! x + 5! x5 = cosx = 2! x2 + 4! x4 = ( ) (2+)! x2+ ( ) (2)! x2 (9) Example 62 Example 6 (e) We have see, i (5), that e x = +x+ x2 2! + + x! + (+)! ec x + (0) for some c betwee 0 ad x. We ca use this to fid approximate values for the umber e, with ay desired degree of accuracy. Just settig x = i (0) gives e = ++ 2! + +! + (+)! ec () for some c betwee 0 ad. Sice e c icreases as c icreases, this says that !! e is a approximate value for e with error at most. The oly problem with this error (+)! boud is that it cotais the umber e, which we do ot kow. Fortuately, we ca agai use () to get a simple upper boud o how big e ca be. Just settig = 2 i (), ad agai usig that e c e, gives e ++ 2! + e! = (! ) ( ) e = 6 e ++ = 5 = e 5 6 = 2! Sequeces ad Series 9 March 8, 206

40 So we ow kow that is a approximate value for e with error at most 2!!. For example, whe = 9, = < (+)! (+)! 0! 0 6 so that with ++ 2! + + 9! e ++ 2! + + 9! ! +! + 4! + 5! + 6! + + 7! + 8! = = ! to six decimal places. Example 6 Example 64 (π) There are umerous methods for computig π to ay desired degree of accuracy. May of them use the Taylor expasio 5 arctax = ( ) x2+ 2+ that we derived i Example 60. Oe of the simplest uses ta π 6 = : ( ) π = 6arcta = 6 = 2 = 2 ( ( ) 2+ ( ) 2+ ( ) ) This is a alteratig series. So, by Theorem, the error we itroduce by trucatig it is betwee zero ad the first term dropped. For example, if we keep te terms, stoppig at = 9, we get π =.459 (to 6 decimal places) with a error betwee zero ad 2 < This is just oe of very may ways to compute π. Aother oe, which still uses (2) but is much more efficiet, is π = 6arcta 5 4arcta 29 This formula was used by Joh Machi i 706 to compute π to 00 decimal digits. Example 64 (2) 5 This is ideed the Taylor expasio, with cetre 0, (i.e. Maclauri expasio) of arctax, eve though we did ot use () to derive it. Sequeces ad Series 40 March 8, 206