LECTURES 5 AND 6: CONVERGENCE TESTS

Transcription

1 LECTURES 5 AND 6: CONVERGENCE TESTS I the previous lecture we saw a very basic test for covergece - the compariso test. Here we develop slightly more sophisticated machiery for determiig the covergece of series, Theorem (Limit comparsio test). Suppose (a ), (b ) are positive sequeces ad a /b c 0. The (a ) is summable if ad oly if (b ) ios summable. Example. Cosider the series log( + / 2 ). Note that log( + x) x 0 x ad so log( + / 2 ) / 2. By the it comparsio test, sice / 2 coverges it follows that log(+ / 2 ) also coverges. The previous example ca be geeralised to show that if (a ) is a positive sequece, the log(+a ) coverges if ad oly if a coverges. For (b ) cosider the sequece of partial products (p ) give by p : b j b b 2... b. j If (p ) coverges, the we write b : p. Suppose b j > 0 for each j N ad ote that log p log b j. j Thus, i this case the study of covergece of ifiite products ca be reduced to the covergece of ifiite series. I particular, from the previous example we see that if (a ) is a positive sequece, the ( + a ) coverges if ad oly if I particular, ot coverge). a coverges. ( + / 2 ) coverges, but ( + /) diverges (i.e. does

2 2 LECTURES 5 AND 6: CONVERGENCE TESTS Theorem (Ratio test). Suppose (a ) is a positive sequece ad a + r. a If r <, the a coverges. If r >, the a diverges. x Example. If 0 < x is ay fixed umber, the coverges.! Suppose 0 < x ad N are fixed. If x < (respectively, x > ), the x coverges (respectively, diverges). If the it ratio satisfies r, the the test is icoclusive. For example, cosider a / ad b / 2. The followig test geeralises our proof of the divergece of the harmoic series. Theorem (Itegral test). Suppose that f : [, ) R is positive, cotiuous ad decreasig ad let a : f() for all N. The a coverges if ad oly if the improper itegral exists. Example. Let p > 0. The series f : R R f / p coverges if ad oly if p >. Theorem (Cauchy codesatio test). For a o-egative, o-icreasig sequece (a ), the series a coverges if ad oly if the codesed series 2 a 2 coverges. 0 Proof. Cosider the sequeces (b ) ad (c ) defied by b : a 2 + if 2 < 2 + for some N {0} c : a 2 if 2 < 2 + for some N {0} Hece, the sequece (b, b 2,... ) is give by (a 2, a 4, a 4, a 8, a 8, a 8, a 8, a 6, a 6, a 6, a 6, a 6, a 6, a 6, a 6,... ) whilst (c, c 2,... ) is give by (a, a 2, a 2, a 4, a 4, a 4, a 4, a 8, a 8, a 8, a 8, a 8, a 8, a 8, a 8,... ). Sice (a ) is o-icreasig it follows that b a c for all N. Hece, by the compariso test, if a coverges, the b coverges. By the defiitio of the b, the latter series coverges if ad oly if 2 a 2 2 a 2 2 coverges.

3 O the other had, if LECTURES 5 AND 6: CONVERGENCE TESTS 3 c coverges, the a coverges. By the defiitio of the c, the former series coverges if ad oly if 2 a 2 coverges. Combiig these observatios completes the proof. Example. Let p > 0. The series log p coverges if ad oly if p >. 2 Hitherto we have oly cosidered o-egative sequeces (a ). Covergece tests for o-positive series are a immediate cosequece of this theory: give (a ) with a 0 for all N we ca simply ote this sequece is summable if ad oly if the o-egative sequece (a ) is summable, i which case a a. Issues of covergece are far more subtle if the terms are allowed to have differet sigs, sice i this case the cacellatio betwee terms ofte becomes crucial to esure covergece. Theorem (Alteratig sig test). Suppose (a ) is o-egative, o-icreasig ad satisfies a 0. The the series () a coverges. Example. Recall that the harmoic series (/) is ot summable. However, the theorem implies that () coverges. Thus, i this example the cacellatio betwee the terms of the sequece is vital to esure covergece. Proof (of the alteratig sig test). The ey ideas of the proof are as follows. Show that: ) s 2+2 s 2 for all N; 2) s 2 s 2+ for all N; 3) s l s for l odd ad eve. By ) ad 3) the subsequece (s 2 ) is o-icreasig ad bouded below ad therefore is coverges to β : if{s 2 : N}. By 2) ad 3) the subsequece (s 2 ) is o-decreasig ad bouded above ad therefore is coverges to α : sup{s 2 : N}. By 3), α β. Sice s 2 s 2 a 2 0 as, it follows that α β. We coclude s α β. Example. I some cases although there is cacellatio, it is t really eeded to esure covergece. For istace, cosider the series () 2 ; 0

4 4 LECTURES 5 AND 6: CONVERGENCE TESTS this coverges by the alteratig series test, but eve if there were t ay () s we d still have covergece - the cacellatio betwee terms is just a added bous! We d lie to idetify the sequeces which are lie the previous example, where the cacellatio is t eeded to esure covergece. Defiitio (Absolutely coverget series). The series a is absolutely coverget if a coverges. Example. () is absolutely coverget. 2 Theorem. Every absolutely coverget series is coverget. Proof. The ey ideas of the proof are: By the Cauchy criterio m+ a ad so (a ) is summable by the Cauchy criterio. a m+ m, m+ a 0. We fiish o a example which shows how difficult it ca be to exploit the cacellatio i certai situatios. Example. Cosider the series si. Although the harmoic series diverges, the oscillatio of the si fuctio should itroduce some cacellatio ad so there s a chace that this series could coverge. Ideed, if we cosider the improper itegral dx (0.) x the we recall that this does ideed coverge. Now we caot use the itegral test for covergece here sice the fuctio /x is ot mootoe. However, the techiques we use to show the covergece of the itegral ca be used to shie a light o how to aalyse the sum. I particular, to show that (0.) coverges oe first uses itegratio by parts to write b x dx cos xb b x x x x 2 dx. Now, sice b b dx b x 2 dx b x 2, by splittig the itegral b ito a two pieces depedig o the sig o cos, oe ca easily show that exists ad so x x 2 dx cos dx x 2 dx.

5 LECTURES 5 AND 6: CONVERGENCE TESTS 5 We d lie to use a similar argumet for sums. For this we use a discrete versio of itegatio by parts, called summatio by parts. I the familiar cotiuous versio, we thi of the itegrad as f g where f is a primitive of si; that is f(x) : x 0 si(t) dt. The discrete aalogue of the primitive is give by replacig the itegral with a sum, thus: si(( + )/2) si(/2) F () : si(j). si(/2) j0 Here the secod equality follows by the double agle formula, as described i exercise 5-33 of the textboo. O the other had, the discrete versio of the derivative is just a differece betwee two cosecutive terms. Sice si F () F ( ), we see F does ideed act lie a primitive. We ow substitute this ito our partial sum to obtai si F () F ( ) F () F () F ( ) F () + 0 F () F (0) + ( ) F (). + This is the discrete aalogue of the itegratio by parts formula. Note that + ( + ) is very close to / 2. Although F () does t loo very much lie cos, it does share the ey property that it is uiformly bouded i : F () si(( + )/2) si(/2) si(/2) si(/2) Thus, from these observatios we coclude that ( ) F () + is absolutely coverget, ad therefore coverget. Furthermore, it follows that si F () + ( ) F () + is coverget (ote that the first it o the right-had side is 0). Joatha Hicma, Departmet of mathematics, Uiversity of Chicago, 5734 S. Uiversity Aveue, Echart hall Room 44, Chicago, Illiois, address: jehicma@uchicago.edu