6.3 Testing Series With Positive Terms


 Marilynn Burns
 4 years ago
 Views:
Transcription
1 6.3. TESTING SERIES WITH POSITIVE TERMS 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 sums S = kow that lim S = a i ad testig it for covergece. We i= a i. So, if the sequece of partial sums coverges, i= so will the series ad they coverge to the same umber. If the sequece of partial sums diverges, so will the series. I theory, this souds like the problem is solved, sice we kow how to study the covergece of sequeces. I practice though it is ot as easy. I geeral, we will ot have a ice formula for S, therefore it may be diffi cult to determie if it coverges. This techique does work i some examples. Whe the sequece of partial sums caot be used, there are several alteratives which we list here: If the sequece is a kow series, use your kowledge of that series. This is why, as you lear ew series, it is importat to remember them. So far, we kow the followig series: Harmoic series: Geometric series: diverges otherwise.. It diverges. i ar i. It coverges to i= i= a r if r < ad If the sequece is ot kow, the first test that should be applied is the test for divergece. It says that if lim 0, the the series a i will diverge. If this limit is zero, the test provides o i= coclusio. Other tests must be used. Remark 6.3. It should be obvious to the reader that covergece of a series is ot affected by the first fiitely may terms of the series. I other words, if coverges, so will a i, i= a i,..., i=3 i= a i where p is ay positive iteger. The i=p same is true for divergece. What will chage however, is the sum. Obviously, if a i
2 308 CHAPTER 6. INFINITE SEQUENCES AND SERIES a i coverges say to L,that is if a i = L, the we also kow that a i L. i= It is easy to fid what it is though. i= a i = L a. (why?) i= The rest of this documet presets some of the tests which ca be used to test series for covergece. We first begi with tests which oly apply to series havig positive terms. Later o we will look at tests for series havig mixed terms. The rest of this sectio is oly cocered with series of positive terms that is series of the form a i where a i 0 for every i. i= 6.3. The Itegral Test This test compares a series to a itegral (improper). It establishes that uder certai coditios, covergece of a series ca be established by studyig the covergece of a improper itegral. It eve goes further. The value of the itegral ca also be used to approximate the ifiite series. However, ulike for geometric series, this test will ot give us a exact value for coverget ifiite series. Theorem 6.3. (Itegral test) If f is a cotiuous, positive ad decreasig fuctio o [, ) ad a = f () the a i coverges if ad oly if f (x) dx i= coverges. Proof. Usig the otatio S = a i with a i = f (i) where f satisfies the i= coditios of the itegral test, it is easy to see that + f (x) dx S a + f (x) dx Sice a i 0, {S } is a icreasig sequece. If the itegral coverges, lim f (x) dx is fiite hece a + f (x) dx is also fiite. Thus {S } is bouded above. Sice it is also icreasig, it must coverge. Hece a i coverges. O the other had, if the itegral diverges, + f (x) dx as. Hece lim S = thus a i diverges. i= Remark The coditios i the itegral test do ot have to hold startig at =, as log as they hold from some poit o. However, if they hold o a iterval of the form [ 0, ) for some positive iteger 0 the oe will study the covergece of the itegral 0 f (x) dx istead of f (x) dx. i= i=
3 6.3. TESTING SERIES WITH POSITIVE TERMS 309 Remark Before applyig this test, make sure that the series satisfies all the coditios of the theorem. Remark The theorem does ot idicate what the series coverges to. You should ot assume that the value of the itegral ad the series are the same. Example Test = l for covergece. The fuctio defiig the geeral term of this series is f (x) = l x x. It is cotiuous ad positive o [, ). To determie if it is decreasig, we study the sig of its derivative. f (x) = l x. Sice the deomiator is a square, x it is always positive. So the sig of the derivative is uiquely determied by its umerator. f (x) < 0 l x < 0 < l x l x > x > e Therefore, f is ot decreasig o [, ) but it is o [3, ), which is good eough. Remember, this property oly eeds to hold from some poit o, which it does here. Havig verified that all the hypotheses of the theorem were satisfied, we ca ow use the itegral test. We study the covergece of a improper itegral. 3 l x x dx = lim t 3 = lim t = t l x x dx [ (l t) The itegral diverges, therefore the series diverges. ] (l 3) 3 l x x dx. This is Remark The above itegral was evaluated usig the substitutio u = l x therefore du = dx x. Example Test i= i p for covergece. If p < 0, the lim i i p =, so the series diverges. If p = 0, the i p = i 0 =.
4 30 CHAPTER 6. INFINITE SEQUENCES AND SERIES hece, lim =. The series also diverges. If p > 0, the the fuctio i ip f (x) = x p is cotiuous, decreasig (sice its deomiator is icreasig) ad positive o [, ). So, istead of studyig the covergece of the series, we study the covergece of the itegral dx. From a theorem about improper xp itegrals, we kow that this itegral coverges if p > ad diverges otherwise that is i this case whe 0 < p. Combiig all the cases, we see that the series coverges whe p > ad diverges otherwise. Remark This proves that the harmoic series, a pseries with p =. i=, diverges sice it is i The above example is very importat. Its result should be remembered. Series which appear i the above example have a special ame. Defiitio A pseries is a series of the form Example 6.3. Here are some examples of pseries. i= i p.. 3. i= i= i= (p = ) i i (p = ) i 3 (p = 3 ) I example 6.3.8, we proved the followig: Theorem 6.3. A pseries coverges if p >. It diverges otherwise. Example Test i i= for covergece. This is a pseries with p =, therefore it coverges. Example Test i= i for covergece. This is a pseries with p =, therefore it diverges.
5 6.3. TESTING SERIES WITH POSITIVE TERMS Compariso Tests The idea here is to compare a ukow series to a kow series. Series such as the harmoic series, geometric series ad pseries are used a lot. The questio is, give a ukow series, how do we kow what to compare it to. Series which look like a pseries or a geometric series should be compared with such series. For example, = +, thus = + should be compared with = +. Similarly, = because for large, should be compared to. I additio, the followig facts are ofte used ad should be remembered. If a, b, c are positive umbers the b > c implies that a b < a c l x < x < e x Oe of two compariso tests ca be used: Theorem (Stadard compariso test) Suppose that a ad b are two series with positive terms such that a b for every from some poit o. The, If b coverges, so does a (i.e. if the series with larger terms coverges, so will the series with smaller terms). If a diverges, so does b (i.e. if the series with smaller terms diverges, so will the series with larger terms). Ituitively this theorem should make sese. Sice we have series of positive terms, either the series is fiite, i which case it coverges or it is ifiite, i which case it diverges. So, if 0 a b the 0 a b So, we see that if the series of larger terms, b coverges that is is fiite, the a must also be fiite hece coverge. Equivaletly, if a diverges, that is is ifiite, the b must also be ifiite that is diverge. These are the oly two situatios uder which we ca coclude. Remark It is easier to remember the theorem i this form: Give two series of positive terms. If the series of larger terms coverges, so does the series of smaller terms. Equivaletly, if the series of smaller terms diverges, so does the series of larger terms. The idea here is that if we suspect that the give series coverges, we eed to fid a series with larger terms which we kow coverges. Coversely, if we
6 3 CHAPTER 6. INFINITE SEQUENCES AND SERIES suspect the give series diverges, we eed to fid a series with smaller terms which we kow diverges. It is also importat to ote that the iequality does ot have to hold for every. It just has to hold from some poit o. Example Test for covergece. + 5 For large, + 5 behaves like. So, we compare + 5 to which is a pseries with p =, thus it coverges. At this poit, we caot coclude yet that the give series coverges. For this, we eed to prove that the terms of. This is easy to see because + 5 are smaller tha the terms of ad therefore follows that + 5 coverges. Example Test l. By the stadard compariso test, it for covergece. Sice l for 3, it follows that l By the stadard compariso test, l Example Test +. We kow that diverges. diverges. for covergece. For large, + behaves like. We also kow that coverges (geometric series with r = ). Thus we suspect our series coverges. Sice +, we ca use the stadard compariso test to coclude that it does. Example Test 5 for covergece. Sice 5 behaves like, we suspect our series coverges sice coverges. However, 5. So, we caot use the stadard compariso test. The ext compariso test addresses this issue. Theorem 6.3. (Limit compariso test) Suppose that a ad b are two series with positive terms such that L = lim. If 0 < L <, the two series behave alike. a b exists or is the:. If L = 0 the covergece of b implies covergece of a. Divergece of a implies divergece of b.
7 6.3. TESTING SERIES WITH POSITIVE TERMS If L = the covergece of a implies covergece of b. Divergece of b implies divergece of a. Ituitively, the theorem should make sese. If lim a b = 0 the this idicates that b is much larger tha a. Thus, b is the series with larger terms. We ca use remark to help us remember what the theorem allows us to a coclude. Similarly, if lim =, the a is much larger tha b. So, this b time a is the series with larger terms. Example 6.3. Test for covergece. 5 This is the series we could ot hadle with the stadard compariso test. Agai, we compare it to. This time, we look at lim 5 = lim = 5 also co Sice, we coclude by the limit compariso test that 5 verges Usig the Itegral Test to Estimate Sums As poited out above, the itegral test ca be used to study the covergece of certai series. But, the itegral ad the series do ot have the same value. However, the itegral ca be used to fid the error whe approximatig the sum of a series by its sequece of partial sums. Let us itroduce some otatio. Let us assume that we have a series a i, which satisfies the coditios of the itegral test. Assume we have prove that this series coverges. We would like to kow what it coverges to. Call S = a i Obviously, we caot add up all the terms; there is a ifiite amout. i= But we kow from the test for divergece that because the series coverges, lim a = 0. Therefore, the terms we are addig become smaller ad smaller. Evetually, they will be so small that they will almost ot cout. This suggests that we could approximate this ifiite sum by addig a fiite umber of terms ad igore the remaiig oes. I other words, we could approximate S = a i by S = a i. Of course, the two will ot be exactly equal. Whe we i= i= approximate somethig with somethig else, it is importat to kow the error we are makig. Let us deote the error by R. I other words, R = S S. i=
8 34 CHAPTER 6. INFINITE SEQUENCES AND SERIES It turs out that ot oly we ca fid the error, we ca also fid how may terms we eed to add if we eed the error to be less tha a certai umber. This is the ext theorem, which we will state without proof. The proof ca be foud i ay calculus book. Theorem If a i coverges ad satisfies the coditios of the itegral i= test ad if we set R = S S, the + f (x) dx R f (x) dx (6.) ad therefore S + f (x) dx S S + f (x) dx (6.) + We illustrate with a few examples how this theorem ca be used to fid the error whe we approximate S with S as well as to fid out how may terms must be added so that the error is less tha a certai amout. The secod idea is extremely importat. We kow that whe we build somethig, whether it is a bridge, or tiy parts for a computer, we ca ever build these parts to the exact measuremets. Hece, there is the idea of "tolerace", that is the error we are allowed for the part to still fuctio. Obviously, the tolerace for a bridge is much larger tha for computer parts. This idea of tolerace exists i pretty much everythig we do. So, it may happe that a real problem has a ifiite series as its solutio. Because we may ot be able to fid exactly what the ifiite series is, we may have to approximate it. The tolerace of the problem we are solvig will tell us the maximum error our approximatio ca have. If you thik about it, this is pretty amazig. We do ot kow what the exact sum of the series is, yet we ca tell how far from the exact value our approximatio is. Example Cosider the series. Suppose we approximate this series with S 0 (the sum of the first 0 terms). What will be the error ad what i3 i= ca we say about the exact value of i 3? i= Let S deote the exact sum. We are approximatig S with S 0. Usig the otatio above, the error is R 0. From the theorem, we see that x 3 dx R 0 0 x 3 dx
9 6.3. TESTING SERIES WITH POSITIVE TERMS 35 We begi by computig the itegrals. Therefore, Ad t dx = lim x3 t x 3 dx [ ] t = lim t x [ = lim t t + ] = x 3 dx = () 0 = 4 = x 3 dx = (0) = 00 =.005 It follows that if we approximate S by S 0, the error R 0 will satisfy R I other words, the error is at least but o greater tha.005. I fact, we ca eve do better, usig the secod iequality of the theorem, ad 0 the fact that S 0 = i 3 = = , we see that i= S 0 + x 3 dx S S x 3 dx S S Thus, if we approximate S by the midpoit of these two values, we have S
10 36 CHAPTER 6. INFINITE SEQUENCES AND SERIES Ad the error is less tha half the differece betwee these two values that is R Remark What the above example suggests is that we ca approximate S by usig the midpoit of the iterval S + + f (x) dx, S + f (x) dx. I doig so, the error is less tha half the width of the same iterval, that is R f (x) dx + f (x) dx This provides a better approximatio as the example below will cofirm. (6.3) Example Approximate the series with a error less tha.00. i i= I other words, we wat to fid (how may term we should add) so that R <.00.. Usig the first approach. We wat to approximate i i= with S, so that R.00. From the theorem ad equatio 6., we kow that R dx. Thus, if we solve x begi by computig the itegral. dx.00, what we wat will follow. We x t dx = lim x t x dx [ ] t = lim t x [ = lim + ] t t To have withi.00 we use =.00, we must have.00 = 000. To approximate i 000 i= i. Ideed, i= i = 6 π = ad i= 000 i= i =
11 6.3. TESTING SERIES WITH POSITIVE TERMS With this approach, we had to add the first 000 terms of the series.. Usig equatio 6.3, we wat to fid so that This happes whe + ( + ) x dx + x dx.00. Usig your calculator, you ca try values of util the iequality is satisfied. Whe =, ( + ) = (4) () = >.00. But whe =, ( + ) = (44) (3) = <.00. This meas that that if we approximate S by the midpoit of S + 3 f (x) dx, S + the error will be less tha.00. You may thik this is more work, but it is ot the case. Most of the work is i addig the terms of the series. Above, we added the first 000 terms. Here, we oly add the first terms. Furthermore, ad 3 S = i= i = f (x) dx = 3 f (x) dx = = = Therefore, the midpoit of the iterval S + 3 f (x) dx, S + f (x) dx f (x) dx,
12 38 CHAPTER 6. INFINITE SEQUENCES AND SERIES is S + f (x) dx + S + f (x) dx 3 = = This is withi the desired accuracy Thigs to Kow Studets should kow ad be able to use the followig tests Itegral test pseries, what they are, whe they coverge Stadard ad limit compariso tests Studets should also be able to approximate the sum of a series, ad determie the error of the approximatio Problems. Suppose that Σa ad Σb are series with positive terms ad Σb is kow to be coverget. (a) If a > b for all, what ca be said about Σa? Why? (b) If a < b for all, what ca be said about Σa? Why?. It is importat to distiguish betwee b ad = b. What ame is give to the first series? To the secod? For what value of b doe the first series coverge? For what value of b doe the secod series coverge? 3. Use the itegral test to determie if the itegrals below coverge or diverge. = (a) (b) = = 4 e 4. Use a compariso test to determie if the series below coverge or diverge. (a) = + +
13 6.3. TESTING SERIES WITH POSITIVE TERMS 39 (b) (c) (d) = = = cos cos 5. Determie whether the series below coverge or diverge usig ay of the tests studied so far. (a) (b) e (c) (d) (e) (f) (g) (h) (i) = = = = l ( ) si + ( ) = = = = + si Fid the values of p for which 7. Cosider the series = 4. = (l ) p is coverget. (a) Use the sum of the first 0 terms to estimate estimate? =.How good is this 4
14 30 CHAPTER 6. INFINITE SEQUENCES AND SERIES (b) How may terms should we add if we wat the error to be less tha 0 6? 8. The meaig of the decimal represetatio of a umber 0.d d d 3 d 4... (where d i is a iteger betwee 0 ad 9) is that 0.d d d 3 d 4... = d 0 + d 0 + d d Show that this series always coverge Aswers. Suppose that Σa ad Σb are series with positive terms ad Σb is kow to be coverget. (a) If a > b for all, what ca be said about Σa? Why? othig. (b) If a < b for all, what ca be said about Σa? Why? Σa coverges by the stadard compariso test.. It is importat to distiguish betwee b ad = b. What ame is give to the first series? To the secod? For what value of b doe the first series coverge? For what value of b doe the secod series coverge? b : p series, coverges for b <. = = b : geometric series, coverges whe b <. = 3. Use the itegral test to determie if the itegrals below coverge or diverge. (a) (b) 4 = coverges. = e coverges. 4. Use a compariso test to determie if the series below coverge or diverge. (a) + + = coverges.
15 6.3. TESTING SERIES WITH POSITIVE TERMS 3 (b) (c) (d) cos 3 = coverges. 3 = coverges. + cos = coverges. 5. Determie whether the series below coverge or diverge usig ay of the tests studied so far. (a) coverges. (b) e = coverges. (c) l = diverges. 5 (d) + 3 = coverges. + (e) = diverges. + (f) 4 + = coverges. ( ) (g) si = diverges. + ( ) (h) = Coverges. + si (i) 5. = coverges.
16 3 CHAPTER 6. INFINITE SEQUENCES AND SERIES 6. Fid the values of p for which coverges if ad oly if p >. 7. Cosider the series = 4. = (l ) p is coverget. (a) Use the sum of the first 0 terms to estimate estimate? R =.How good is this 4 (b) How may terms should we add if we wat the error to be less tha 0 6? 70 terms. 8. The meaig of the decimal represetatio of a umber 0.d d d 3 d 4... (where d i is a iteger betwee 0 ad 9) is that 0.d d d 3 d 4... = d 0 + d 0 + d d Show that this series always coverge. hit: compare it to a geometric 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 informationMath 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 informationINFINITE 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 informationINFINITE 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 informationMath 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 informationMAT1026 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 information10.6 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 informationChapter 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 informationPart 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 informationChapter 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 informationTesting 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 informationSeries 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 information7 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 informationChapter 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 informationSCORE. Exam 2. MA 114 Exam 2 Fall 2016
MA 4 Exam Fall 06 Exam Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use
More informationChapter 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 informationRoberto 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 informationIn this section, we show how to use the integral test to decide whether a series
Itegral Test Itegral Test Example Itegral Test Example pseries Compariso Test Example Example 2 Example 3 Example 4 Example 5 Exa Itegral Test I this sectio, we show how to use the itegral test to decide
More informationChapter 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 information4x 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 informationMath 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 informationSequences. 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 informationCarleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of truefalse questions, like the one below.
Carleto College, Witer 207 Math 2, Practice Fial Prof. Joes Note: the exam will have a sectio of truefalse questios, like the oe below.. True or False. Briefly explai your aswer. A icorrectly justified
More informationCHAPTER 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 informationSCORE. Exam 2. MA 114 Exam 2 Fall 2016
Exam 2 Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use a graphig calculator
More information9.3 The INTEGRAL TEST; pseries
Lecture 9.3 & 9.4 Math 0B Nguye of 6 Istructor s Versio 9.3 The INTEGRAL TEST; pseries I this ad the followig sectio, you will study several covergece tests that apply to series with positive terms. Note
More informationsin(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 informationSection 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 informationMath 132, Fall 2009 Exam 2: Solutions
Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of
More informationAn alternating series is a series where the signs alternate. Generally (but not always) there is a factor of the form ( 1) n + 1
Calculus II  Problem Solvig Drill 20: Alteratig Series, Ratio ad Root Tests Questio No. of 0 Istructios: () Read the problem ad aswer choices carefully (2) Work the problems o paper as eeded (3) Pick
More informationMath 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 informationMa 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 informationPhysics 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 informationMa 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 informationSeries 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 informationA 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 informationReview for Test 3 Math 1552, Integral Calculus Sections 8.8,
Review for Test 3 Math 55, Itegral Calculus Sectios 8.8, 0.0.5. Termiology review: complete the followig statemets. (a) A geometric series has the geeral form k=0 rk.theseriescovergeswhe r is less tha
More informationPractice 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 information1 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 informationSequences 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(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 information5.6 Absolute Convergence and The Ratio and Root Tests
5.6 Absolute Covergece ad The Ratio ad Root Tests Bria E. Veitch 5.6 Absolute Covergece ad The Ratio ad Root Tests Recall from our previous sectio that diverged but ( ) coverged. Both of these sequeces
More informationE. Incorrect! Plug n = 1, 2, 3, & 4 into the general term formula. n =, then the first four terms are found by
Calculus II  Problem Solvig Drill 8: Sequeces, Series, ad Covergece Questio No. of 0. Fid the first four terms of the sequece whose geeral term is give by a ( ) : Questio #0 (A) (B) (C) (D) (E) 8,,, 4
More informationSUMMARY 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 informationSequences 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 informationZ 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 informationTHE INTEGRAL TEST AND ESTIMATES OF SUMS
THE INTEGRAL TEST AND ESTIMATES OF SUMS. Itroductio Determiig the exact sum of a series is i geeral ot a easy task. I the case of the geometric series ad the telescoig series it was ossible to fid a simle
More informationSeunghee 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 informationAP 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 informationSection 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 informationSOLUTIONS 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 informationInfinite 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 informationMA131  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 informationAlternating 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 informationPlease do NOT write in this box. Multiple Choice. Total
Istructor: Math 0560, Worksheet Alteratig Series Jauary, 3000 For realistic exam practice solve these problems without lookig at your book ad without usig a calculator. Multiple choice questios should
More informationSequences, 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 informationJANE PROFESSOR WW Prob Lib1 Summer 2000
JANE PROFESSOR WW Prob Lib Summer 000 Sample WeBWorK problems. WeBWorK assigmet Series6CompTests due /6/06 at :00 AM..( pt) Test each of the followig series for covergece by either the Compariso Test or
More informationMA131  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 informationPLEASE 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 informationENGI Series Page 601
ENGI 3425 6 Series Page 601 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 informationMATH 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 informationNotice that this test does not say anything about divergence of an alternating series.
MATH 572H Sprig 20 Worksheet 7 Topics: absolute ad coditioal covergece; power series. Defiitio. A series b is called absolutely coverget if the series b is coverget. If the series b coverges, while b diverges,
More informationMath 106 Fall 2014 Exam 3.2 December 10, 2014
Math 06 Fall 04 Exam 3 December 0, 04 Determie if the series is coverget or diverget by makig a compariso (DCT or LCT) with a suitable b Fill i the blaks with your aswer For Coverget or Diverget write
More informationMTH 246 TEST 3 April 4, 2014
MTH 26 TEST April, 20 (PLEASE PRINT YOUR NAME!!) Name:. (6 poits each) Evaluate lim! a for the give sequece fa g. (a) a = 2 2 5 2 5 (b) a = 2 7 2. (6 poits) Fid the sum of the telescopig series p p 2.
More information2 n = n=1 a n is convergent and we let. i=1
Lecture 3 : Series So far our defiitio of a sum of umbers applies oly to addig a fiite set of umbers. We ca exted this to a defiitio of a sum of a ifiite set of umbers i much the same way as we exteded
More informationf(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 information62. 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 informationn=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n
Series. Defiitios ad first properties A series is a ifiite sum a + a + a +..., deoted i short by a. The sequece of partial sums of the series a is the sequece s ) defied by s = a k = a +... + a,. k= Defiitio
More information1 Introduction to Sequences and Series, Part V
MTH 22 Calculus II Essex Couty College Divisio of Mathematics ad Physics Lecture Notes #8 Sakai Web Project Material Itroductio to Sequeces ad Series, Part V. The compariso test that we used prior, relies
More informationSection 5.5. Infinite Series: The Ratio Test
Differece Equatios to Differetial Equatios Sectio 5.5 Ifiite Series: The Ratio Test I the last sectio we saw that we could demostrate the covergece of a series a, where a 0 for all, by showig that a approaches
More informationMath 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 informationMIDTERM 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 informationMath 112 Fall 2018 Lab 8
Ma Fall 08 Lab 8 Sums of Coverget Series I Itroductio Ofte e fuctios used i e scieces are defied as ifiite series Determiig e covergece or divergece of a series becomes importat ad it is helpful if e sum
More information1 Lecture 2: Sequence, Series and power series (8/14/2012)
Summer JumpStart Program for Aalysis, 202 SogYig Li Lecture 2: Sequece, Series ad power series (8/4/202). More o sequeces Example.. Let {x } ad {y } be two bouded sequeces. Show lim sup (x + y ) lim
More informationThe 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 informationSCORE. Exam 2. MA 114 Exam 2 Fall 2017
Exam Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use a graphig calculator
More informationSeries: Infinite Sums
Series: Ifiite Sums Series are a way to mae sese of certai types of ifiitely log sums. We will eed to be able to do this if we are to attai our goal of approximatig trascedetal fuctios by usig ifiite degree
More information10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1
0. Positive Term Series: Compariso Tests Cotemporary Calculus 0. POSITIVE TERM SERIES: COMPARISON TESTS This sectio discusses how to determie whether some series coverge or diverge by comparig them with
More informationSequences 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 informations = and t = with C ij = A i B j F. (i) Note that cs = M and so ca i µ(a i ) I E (cs) = = c a i µ(a i ) = ci E (s). (ii) Note that s + t = M and so
3 From the otes we see that the parts of Theorem 4. that cocer us are: Let s ad t be two simple oegative Fmeasurable fuctios o X, F, µ ad E, F F. The i I E cs ci E s for all c R, ii I E s + t I E s +
More informationMath 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 informationRead 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 informationCalculus BC and BCD Drill on Sequences and Series!!! By Susan E. Cantey Walnut Hills H.S. 2006
Calculus BC ad BCD Drill o Sequeces ad Series!!! By Susa E. Catey Walut Hills H.S. 2006 Sequeces ad Series I m goig to ask you questios about sequeces ad series ad drill you o some thigs that eed to be
More informationMA131  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(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 informationSection 1.4. Power Series
Sectio.4. Power Series De itio. The fuctio de ed by f (x) (x a) () c 0 + c (x a) + c 2 (x a) 2 + c (x a) + ::: is called a power series cetered at x a with coe ciet sequece f g :The domai of this fuctio
More informationMath 106 Fall 2014 Exam 3.1 December 10, 2014
Math 06 Fall 0 Exam 3 December 0, 0 Determie if the series is coverget or diverget by makig a compariso DCT or LCT) with a suitable b Fill i the blaks with your aswer For Coverget or Diverget write Coverget
More informationMath 116 Practice for Exam 3
Math 6 Practice for Exam Geerated October 0, 207 Name: SOLUTIONS Istructor: Sectio Number:. This exam has 7 questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur
More informationMath 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 informationMTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1
MTH 42 Exam 3 Spr 20 Practice Problem Solutios No calculators will be permitted at the exam. 3. A pigpog ball is lauched straight up, rises to a height of 5 feet, the falls back to the lauch poit ad bouces
More informationMath 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 informationMidterm Exam #2. Please staple this cover and honor pledge atop your solutions.
Math 50B Itegral Calculus April, 07 Midterm Exam # Name: Aswer Key David Arold Istructios. (00 poits) This exam is ope otes, ope book. This icludes ay supplemetary texts or olie documets. You are ot allowed
More informationChapter 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 informationConvergence 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 informationLesson 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 informationArkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan
Arkasas Tech Uiversity MATH 94: Calculus II Dr Marcel B Fia 85 Power Series Let {a } =0 be a sequece of umbers The a power series about x = a is a series of the form a (x a) = a 0 + a (x a) + a (x a) +
More informationBC: Q401.CH9A Convergent and Divergent Series (LESSON 1)
BC: Q40.CH9A Coverget ad Diverget Series (LESSON ) INTRODUCTION Sequece Notatio: a, a 3, a,, a, Defiitio: A sequece is a fuctio f whose domai is the set of positive itegers. Defiitio: A ifiite series (or
More informationAdditional Notes on Power Series
Additioal Notes o Power Series Mauela Girotti MATH 370 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 informationCalculus 2  D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.)
Calculus  D Yue Fial Eam Review (Versio //7 Please report ay possible typos) NOTE: The review otes are oly o topics ot covered o previous eams See previous review sheets for summary of previous topics
More informationMath 163 REVIEW EXAM 3: SOLUTIONS
Math 63 REVIEW EXAM 3: SOLUTIONS These otes do ot iclude solutios to the Cocept Check o p8. They also do t cotai complete solutios to the TrueFalse problems o those pages. Please go over these problems
More information