Solutions to Tutorial 5 (Week 6)
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1 The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial 5 (Wee 6 MATH2962: Real ad Complex Aalysis (Advaced Semester, 206 Web Page: Lecturer: Florica C. Cîrstea Questios mared with * are more difficult questios. Material covered ( Covergece tests for series Outcomes This tutorial helps you to ( ow ad be able to apply covergece tests for sequeces ad series Summary of essetial material Series are sequeces formed by summig terms a K N. We deote series by a. The sequece (s defied by s : for all N is called the sequece of partial sums We say that the series a is coverget if the sequece of partial sums (s coverges, ad it is diverget if the sequece of partial sums (s diverges. Covergece ad divergece of series is determied by applyig covergece tests. Cauchy criterio Applies to: All series Useful mostly for theoretical purposes, rarely used o specific series. For every ε > 0 there exists 0 N so that < ε for all > 0. Compariso ad limit compariso tests Applies to: Series with o-egative terms. a Useful for series i R ad for testig for absolute covergece (apply to a Requires owledge about elemetary series: geometric series, p-series, harmoic series, expoetial series. Suppose that a 0 ad b > 0 for all N. Let oe of the two coditios be satisfied: (i there exists m N such that a b for all m (Compariso test (ii a b is a bouded sequece (Limit Compariso test If b coverges, the a coverges. If a diverges, the b diverges.. Copyright c 206 The Uiversity of Sydey
2 Leibiz test Applies to: alteratig series oly. Useful for alteratig series oly, does ot guaratee absolute covergece. Suppose that a 0 ad a + a for all N, ad that a 0. The ( a coverges. Cauchy codesatio test Applies to: series with o-egative ad decreasig terms. Useful for series where the root ad ratio test is icoclusive Suppose that a 0 ad a + a for all N. The a coverges if ad oly if 2 a 2 coverges. Root ad ratio tests Applies to: series i R N or C N. Useful for testig for absolute covergece Defie L by oe of the followig limits: a + (i L : lim (Ratio test a (ii L : lim sup a (Root test The series a coverges absolutely if L <, diverges if L >, ad o coclusio ca be made if L. The table below shows to wich classes of series a the tests are applicable i terms of properties of a ad the partial sums (s. meas that the test is applicable, meas that the test is ot i geeral applicable. boudedess of (s Compariso test Limit compariso test Cauchy codesatio test Itegral test Leibiz test Ratio test Root test Cauchy criterio a 0 a 0 ad a decreasig a ( a, a decreasig a R (arbitrary sigs a R N or C N a R N or C N, test o a 2
3 Questios to complete durig the tutorial. Classify the give series as either absolutely coverget, coverget, or diverget. (a ( ; 0 Solutio: The series is of the form 0 ( a with a > 0. This is a alteratig series with a 0 as. To chec for covergece we estimate the terms a. Note that I particular a 0 as, but evertheless, by compariso with the harmoic series we see that the give series does ot coverge absolutely. To chec for covergece we use the Leibiz. The oly coditio to chec is whether a is decreasig, at least for all N sufficietly large. We already ow that a 0 as. To chec (a is decreasig, ote that a + a ( ( (b (c (d for all. Hece (a is a positive decreasig sequece with limit zero. By the Leibiz test, the series ( coverges (but ot absolutely ( ; 0 Solutio: Note that /2 3 as. Sice the series is 3/2 coverget (a p-series with p 3/2 >, it follows from the limit compariso test that the give series is absolutely coverget, ad therefore also coverget. 2 ( + 4 ; 0 2 ( Solutio: Note that for all N. Hece, the series coverges by compariso with the geometric series (. 2 This 0 0 shows that the origial series coverges absolutely. (! ; Solutio: We test for absolute coverges by usig the ratio test. We have that ( + -st term -th term ( +! ( + +! ( + / e < as, where we used the elemetary limit ( + / e as. Hece The series coverges absolutely by the ratio test. 3
4 (e (f 0 ( 3 2 ; Solutio: We test for absolute covergece usig the Ratio Test: ( + -st term -th term as. Hece the series coverges absolutely. ( (log. 2 ( < Solutio: To test for absolute covergece ote that log 2 for all > e 2. Hece, 0 (log 2 for all > e2, ad therefore the series (log 2 coverges by compariso with the positive coverget geometric series 2. 0 Hece, the origial series coverges absolutely. 2. Use the Cauchy codesatio test to determie for which p > 0 the followig series coverge. (a (log ; p (b 2 Solutio: Accordig to the Cauchy codesatio test we cosider the covergece of the series 2 2 (log 2 p p (log 2 p (log 2 p. p 2 2 By a result from lectures the latter series, ad hece the origial series, coverges if p > ad diverges if p (0, ]. log (log(log ; p 3 Solutio: We agai use the Cauchy codesatio test ad observe that it is sufficiet to prove the covergece properties of 3 Sice 2 2 log 2 ( log(log 2 p 3 log 2 ( log( log 2 p (log 2 (log p ( ( log( log 2 log p ( p log + log(log log(log 2 log ( log( log 2 p. p as, it follows that 3 coverges if ad oly if log (log(log p 3 coverges. Hece by usig part (a the series coverges if p > (log p ad diverges if p (0, ]. 4
5 (c 2 (log p Solutio: We agai use the Cauchy codesatio test ad observe that it is sufficiet to prove the covergece properties of 2 (log 2 2 p p (log 2 2 p (log 2 p, p 2 2 which diverges for all p > 0 sice 2 p for all p > Let (a ad (b be sequeces i C ad let s : j0 a j. (a Show that a j b j b + s s (b + b. ( j0 Hit: Use the defiitio of s ad write the sum o the right had side as a double sum. The iterchage the order of summatio. Solutio: By iterchagig the order of summatio we see that s (b + b j0 a j (b + b j0 j a j (b + b. To fid the limits for the summatio it is useful to draw the regio i the plae over which we tae the sum: j The vertical lies represet summatio over j first ad the over, the dashed horizotal lies the other way roud. The above formula implies that s (b + b j0 Rearragig ( follows. j0 j a j a j (b + b (b + b j a j (b + b j b + s j0 a j b j. (b Suppose that (s is bouded, that (b + b coverges absolutely, ad that b 0 as. Prove that the series a jb j coverges. This is ow as Dirichlet s test. Solutio: The idea is to use ( ad represet the partial sums i the form a j b j b + s s (b + b (2 j0 5 j0
6 for all N. The we show that the right had side coverges as. Sice b 0 as ad (s is a bouded sequece, we coclude that b + s 0 as. As the sequece (s is bouded there exists a positive costat M such that s M for all N. Moreover, usig that (b + b coverges absolutely, we deduce that s b + b M b + b <. Therefore, s (b + b is absolutely coverget ad hece, coverget. Therefore, as, the right had side of (2 coverges, which meas that j0 a jb j coverges. (c How does the above Dirichlet test geeralise the Leibiz test for alteratig series? Solutio: I the Leibiz criterio we set a : (. The the sequece of partial sums s ( is bouded. I fact s for all N. Assumig that (b decreases to zero we get s (b + b (b b + b 0 b + b 0 ad hece the series s (b + b coverges absolutely. The above is more geeral sice we do ot require the series to alterate, but oly eed that (s be bouded if (b is decreasig to zero. (d Suppose that the sequece of partial sums of the series a is bouded. Show that a coverges. Solutio: I the Dirichlet test we use b. The b + b ( + + for all. Hece, the series b + b coverges absolutely. Moreover, a / 0 as.therefore, Dirichlet s test implies the covergece of. 6
7 Extra questios for further practice 4. (a Let a > 0. By usig partial fractios, fid a expressio for the partial sums of the series (a + (a + +. The fid its limit. Solutio: Usig partial fractios, for every N, the geeral term i the series ca be writte i the form (a + (a + + a + a + +. Hece, for all N, we have s ( (a + (a + + a + a + + a a + + a + a a + + a a + +. Except for the first ad the last term, all terms cacel. Such a series is called a telescopig series. The sequece (s of -th partial sums is thus give by s (a + (a + + a (3 a + + for all N. Lettig, we coclude that (a + (a + + lim s a. *(b Let a > 0. Use partial fractios to write the partial sums of the series (a + (a + + (a i terms of the partial sums i part (a. The fid its limit. Solutio: As i the previous part we get a explicit expressio for the partial sums of the series, ad the compute its limit. Usig partial fractios we ca write the geeral term i the series as (a + (a + + (a ( 2(a + + a + a ( (a + (a + + (a + + (a Hece the -th partial sum of the give series is t (a + (a + + (a ( 2 (a + (a + + (a + + (a (a + (a (a + + (a
8 The two sums o the right had side are partial sums of the same form as those i part (a with a ad a +, respectively. Hece, (a + (a + + (a ( 2 a + ( a a + 2 a (a + (a + 2 2(a + + (a for all 2. Lettig we get (a + (a + + (a (a + (a + 2. Remar. Usig a iductio we could obtai a more geeral result: If a > 0 ad m N \ {0}, the m j0 (a + + j m m (4 j0 (a + j. Ideed, by iductio, it ca be show that the -th partial sum of this series is m j0 (a + + j m m j0 (a + j m m j (a + + j for all. Lettig, we obtai (4. 5. The purpose of this questio is to show that if the terms chage sig, the limit compariso test does ot ecessarily wor. (a Show that the series ( + coverges. Solutio: The series coverges by the Leibiz test sice the series is alteratig ad / decreases to zero. ( (b Show that the series ( + diverges. Solutio: Note that ( + ( ( ( + + ( + + The first of the partial sums coverges by the Leibiz test (see the previous part, but the secod is the harmoic series which diverges to ifiity. Hece the origial series diverges to ifiity. ( (c For, set a : ( + ad b : ( +. Show that a b has a fiite limit as, but the limit compariso test does ot apply.. 8
9 Solutio: We have a b ( ( as. The limit compariso test would imply that the covergece of b implies the covergece of a. However, as show above the series a diverges eve though a coverges. Hece the limit compariso test does ot ecessarily wor if the terms have both sigs. (It wors if the series are absolutely coverget. Challege questios (optioal 6. Let A [a ij ] K N N be a N N matrix. Defie a matrix orm by ( N A : i N /2. a ij 2 Note that this is simply the usual orm i K N N K N 2, ad hece has all its properties icludig the triagle iequality. Prove that the matrix expoetial j e A :! A coverges absolutely. Hit: Use that A A by Tutorial, Questio 9. Solutio: for all. Sice By the hit we have A!! A exp( A! A coverges, it follows that the matrix expoetial coverges absolutely for every matrix A. 9
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