CONVERGENCE TESTS FOR SERIES: COMMENTS AND PROOFS PART I: THE COMPARISON AND INTEGRAL TESTS. Math 112

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1 CONVERGENCE TESTS FOR SERIES: COMMENTS AND PROOFS PART I: THE COMPARISON AND INTEGRAL TESTS Math 112 The covergece tests for series have ice ituitive reasos why they wor, ad these are fairly easy to tur ito rigorous proofs. I these otes we will prove the stadard covergece tests ad give two tests that are't i our text. It's importat to remember that the covergece or divergece of a series depeds oly o what happes to some tail of the series the iclusio or omissio of a ite umber of terms caot chage a coverget series ito a diverget oe or vice versa. (If the series coverges, the sum is a ected, of course.) Because of this, we ca let a little coveiet sloppiess ito our otatio. Whe it does't matter what the startig poit is for a series, 1 we ca write a istead of =1a (or =5a or a ). More speci cally, the =0 P otatio awill always mea a i ite sum, ad will oly be used whe the startig poit for the sum is ot importat. Slight Geeralizatios of the Compariso ad Itegral Tests The observatio above allows the use of the compariso test i cases where the terms of the two series satisfy the appropriate iequality after some poit, but perhaps ot for the rst several terms. Here is the slightly more geeral compariso test. Theorem (Compariso Test). Let a ad b be series. Suppose there is a idex such that 0 a b for all. If b coverges, so does a. Similarly, if a diverges, so does b, ad both sums are i ite. The itegral test has a similar modi catio. 1

2 P Theorem (Itegral Test). Let a be a series. Suppose that is a positive iteger ad thata: [;1)!R is a cotiuous, decreasig fuctio such that a =a() for each P iteger. The either the series a ad the improper itegral a(x)dx both coverge, or they both diverge. The poit here is that the terms of the series ad the fuctio do't have to be related for x ad less tha. I fact the fuctio does't have to be cotiuous, decreasig, or eve de ed. Note that for > 1, if the itegral the series =1 a(x)dx coverges, we ca coclude that a coverges (assumig the idex for the series begis with 1). However, if the series coverges, we caot coclude that the itegral additioal iformatio about the fuctio a. 1 a(x) dx coverges without The Compariso Test. The compariso test is pretty ituitive. Ifa b for all, the it should be the case that =a =b. If =b is ite, so is =a. Similarly, if a is i ite, so is b. It's ot hard to mae this rigorous. = = P P Proof of the Compariso Test. For, we have the iequality =a =b be- P P twee the partial sums. Taig the limit as!1 we have lim =a lim =b!1!1 (which is the same as a b ) provided that the limits exist (the limit1is = = allowed). These limits exist due to the followig lemma about icreasig sequeces. Lemma 1. Supposec, = 1; 2; 3;:::, is a icreasig sequece. If there is some umber L such thatc L for all, the lim c exists as a ite umber. Otherwise lim c =1.!1!1 There are two ways to thi about this iformally. (1) If a icreasig sequece is bouded above, it has to coverge (meaig that the limit is ite). (2) A icreasig sequece has to have a limit (which may be i ite). I the proof of the compariso this lemma is applied to the partial sums A = P = P = adb = b, which are icreasig sice the terms of these sums are o-egative. By the lemma, the limits lim A = a ad lim B = b exist (they may = =!1!1 2 a

3 be i ite), ad the the reasoig precedig the lemma may be applied. The lemma is pretty ituitive also, but its proof uses oe of the fudametal properties of the real umbers, ad so it's worth looig at. (You ca sip the rest of this sectio if you're ot iterested.) Proof of the lemma. Let's tae care of the easy part rst. Suppose that the sequece is ot bouded above. This meas that for every umber L, o matter how large, some term of the sequece is larger tha L, that is, there is some idex N such that c >L. Sice N the sequece is icreasig, we have c c N > L for all N. Thus, for every umber L, there is a idex N such thatc >L for all N. This is exactly what lim c =1 meas.!1 Now suppose that c Lfor all, that is L is a upper boud for the sequece. Now L is ot the oly upper boud ay umber greater thalwill also be a upper boud, ad some umbers less tha L may also be upper bouds. Of these upper bouds, oe of them will be the smallest. Let ` be the least upper boud. I claim that ` is the limit of the sequece. To see this, suppose that a <`. Sice` is the least upper boud of the sequece, the a is ot a upper boud. This meas that some term of the sequece is bigger thaa, that isc >a for some idexn. Sice the sequece is icreasig we have N thata<c N c ` for every N. Thus for ay a<`there is a itegern such that a<c ` for all N. This is (essetially) what lim c =`meas.!1 I geeral, lim c = ` meas that for ay ope iterval cotaiig `, the sequece!1 evetually gets iside the iterval, ad stays iside. More precisely, if a ad b are ay umbers such that a < ` < b, the there is a idex N such that a < c < b for all itegers N. (You may have leared this as \for ay ² > 0 there is a idex N such that` ²<c <`+², orjc `j<², for all N." This just maes the iterval aroud ` symmetric, but that's ot importat.) I the lemma we oly eed to deal with umbers less tha `, sice`is a upper boud for the sequece. The fudametal property of the real umbers used i the lemma is the least upper 3

4 boud property. Least Upper Boud Property. A o-empty set of real umbers that is bouded above has a least upper boud. This property distiguishes the real umbers from the ratioal umbers. It is crucial to may theorems i calculus ad its geeralizatio, aalysis. A study of the least upper boud property ad its sometimes subtle uses are part of the cotet of Math 33 ad 34. Extra Credit Challege Problems. (1) Give a example of a o-empty set of ratioal umbers that is bouded above but does ot have a ratioal least upper boud. (2) Fid some calculus theorems that are't true if you restrict yourself to ratioal umbers. The Itegral Test. The itegral test is also pretty clear oce you see the right pictures (see page 578 of OZ). These suggest that X1 Z 1 X1 a a(x)dx a : =+1 = P1 R1 R1 = As with the compariso test, if a is ite, so is a(x) dx. Similarly, if a(x) dx is ite, so is itervals. = a. A rigorous proof ivolves partial sums ad itegrals over ite Proof of the Itegral Test. For x + 1 we have a a(x) a, sice both +1 the fuctio ad the sequece are decreasig. Itegratig from to + 1 yields a +1 R+1 R a(x)dx a. If we let a = a+1 ad b = a(x)dx, we have a b a. Note that a0 ad a are really the same series, their idices are just o set by oe. It the follows from the regular compariso test that a ad b either both coverge or diverge. P1 R1 Now what is = b? It should be a(x) dx. Ideed, Z 1 Z 1 X Z +1 1 X X1 a(x) dx = lim a(x) dx = lim a(x) dx = lim b = b :!1!1!1 = = = 4

5 R!1 RX There is oe subtlety, however: a(x)dx is't de ed as lim B, where B = a(x)dx; it's de ed as lim B(X), where B(X) = a(x)dx. The di erece is that X!1 is a iteger whereas X is a real umber. These are slightly di eret types of limits. They are obviously related, however, ad i this case they are equal. This follows from two lemmas, the secod of which is the aalogue for fuctios of the lemma i the previous sectio. Lemma 2. Suppose that f : [0;1)!Rad that c = f() for all itegers 0. If lim f(x) exists, the so does lim c, ad they are equal (they may both be i ite). x!1!1 Lemma 3. Suppose that f : [0;1)! R is icreasig. If there is some umber L such thatf(x) L for allx, the limf(x) exists as a ite umber. Otherwise lim f(x) =1. x!1!1 Extra Credit Challege Problems. I problems 5, 6, ad 7 explicit formulas are ot ecessary. Good descriptios are adequate. R1 P1 (3) The itegral test says that a(x) dx coverges if ad oly if a coverges. As with ay \if ad oly if" statemet, there are really two implicatios. Carefully explai how the two lemmas are used i each of these implicatios. I particular, both lemmas are eeded for oe directio, but oly oe is eeded for the other. (4) Give a careful proof of the two lemmas. (5) Give a example of a sequece a, = 0; 1; 2;:::, ad a cotiuous fuctio a: [0;1)!Rwith a =a() for each iteger 0 such that lima exists, but =!1 =0 lima(x) does ot. x!1 P (6) Give a example of a series 1 a of o-egative terms ad a cotiuous fuc tio a: [0;1)! R with a = a() for each iteger 0 such that the series coverges but the improper itegral =0 0 a(x) dx does ot. (7) Give a example of a series a of o-egative terms ad a cotiuous fuc tio a: [0;1)!Rwith a = a() for each iteger 0 such that the improper itegral 0 Robert Foote, October 1999 a(x) dx coverges but the series does ot. 5