A GUIDE TO THE LIMIT COMPARISON TEST

1 A GUIDE TO THELIMITCOMPARISONTESTJ oseph BreenThe LIMIT COMPARISON test is the GOAT infinite series convergence test, but knowing when and how to useit effectively can be difficult. This GUIDE explains the intuition, subtleties, and heuristics of the test andhopefully provides enough elucidating examples. The blue links below are hyperlinks to each The statement of the LIMIT COMPARISON test12 When and how to use LIMIT quick summary .. I: Isolating obvious dominant behavior .. II: Pesky logarithms .. III: Trig functions and other oddities (Taylor polynomials) .. IV: A last ditch resort ..63 Some examples71 The statement of the LIMIT COMPARISON testIn order to use LIMIT COMPARISON , we have to know the statement. I ll provide the mathematical statement,but also how you shouldthinkabout the ( LIMIT COMPARISON test.).Let n=1anbe an infinite series withan>0. Letbn>0be apositive sequence.(i)Iflimn anbnis a finite, positive number, then n=1anand n=1bneither both converge ordiverge.

2 (ii) Iflimn anbn= 0, then if n=1bnconverges, then n=1analso converges.(iii) Iflimn anbn= , then if n=1bndiverges, then n=1analso onlyworks withpositiveseries an bn an bn an' bnWARNING: One thing to keep in mind about cases (ii) and (iii) above is thatyou need to be careful ifyou do LIMIT COMPARISON and you get a LIMIT of0or . In particular, let s say you pick a series bntocompare with, and you know that bndiverges. If you compute thatlimn anbn= 0, you have arguedthat an bn. Since bndiverges, this doesn t tell you anything! Similarly, if you pick a series bnto compare with, you that bnconverges, and you compute thatlimn anbn= , you have argued that an bnand this doesn t tell you anything!12 When and how to use LIMIT comparisonIn my mind, there arethree main scenariosin which LIMIT COMPARISON is helpful, and then a fourth catch-all scenario. The first subsection gives a summary, and the following subsections below describe thesescenarios in A quick summaryIf you are given a (I).

3 Of the form n=1[dominant term] + [fluff][dominant term] + [fluff]then run LIMIT COMPARISON against the series n=1[dominant term][dominant term].Typically this will be ap-series. Also, if you do this correctly your LIMIT computation should giveyoulimn anbn= 1and thus you are in scenario (i) of the LIMIT COMPARISON test.(II) ..with alnnthatyou can t ignore, then generate abnto compare with by replacinglnnwithnsmall #.The small number depends on what else is in the you do this you will almost cer-tainly be in scenario (ii) or (iii) of the LIMIT COMPARISON test, so you need to be careful about nsmall #(III) ..with something likesin(1n), generate abnto compare with by replacing thesin(1n)term are other useful approximations (based on Taylor polynomials) but in 31B,sin(1n)is the mostlylikely (1n) 1n(IV) ..and none of the above situations above apply. If you think that the series diverges, just run limitcomparison against1n. If you think the series converges, run LIMIT COMPARISON against1n2.

4 If you reunsure, just try both! Case I: Isolating obvious dominant behaviorThis is the main situation in which LIMIT COMPARISON applies, and it is also the most series that exhibits obvious dominant terms in the numerator and denominator, you can generate abnto compare with by isolating this dominant behavior. This is best explained with an example: considerthe series n=1n2+ 1 + sinn n7+n5+ first thought when faced with the above series is that there is a lot of fluff that I can ignore. For example,in the numerator, there are three different terms:n2,1, andsinn. Asngrows to infinity, the most importantterm isn2. (The1term does not grow at all, andsin(n)dances around in the interval[ 1,1]. Note that we arenotin Case III described above becausesin(n)ismuchdifferent thansin(1n).) Likewise, in the denominator2there are three termsunder a square root:n7,n5, and1. Thus, the dominant behavior in the denominator is n7. All of this leads me to the following intuition: n=1n2+ 1 + sinn n7+n5+ 1 n=1n2 n7= n=1n2n72= n= n=11n32is a convergentp-series, this intuition suggests that n=1n2+1+sinn n7+n5+1should also LIMIT COMPARISON test is the way to formalize this intuition!

5 Indeed,limn anbn= limn n2+1+sinn n7+n5+1n2 n7= limn 1 +1n2+sinnn2 1 +1n5+1n7=1 + 0 + 0 1 + 0 + 0= a finite, positive number, we are in case (i) of the LIMIT COMPARISON test: n=1n2+1+sinn n7+n5+1and n=11n32either both converge or both diverge. Because the latter series converges, we have concluded bylimit COMPARISON that n=1n2+1+sinn n7+n5+1converges as well! Case II: Pesky logarithmsLogarithms introduce some subtleties in LIMIT COMPARISON . The main takeaway from this section is thefollowing fact:lnn nafor anya > fact,(lnn)p nafor anya >0and anyp > of this means that we can absorb the growth of a logarithm with a very small power ofn. In particular,if there is a logarithm that you can t ignore in a series, you should generate a candidate for comparisonby replacing thelnnterm with a very small power ofn. By doing this, you absorb the logarithmic growthand will likely be in scenario (ii) or (iii) of the LIMIT COMPARISON s easiest to explain this through an example.

6 Consider the following series: n= the denominator grows liken2and the denominator grows much slower thann2, my intuitionsuggests that this series should converge. Let s formalize this with LIMIT : In a situation like this, youcannotjust ignore thelnnand run COMPARISON against1n2. It willfail! Let s check:limn lnnn21n2= limn lnn= . lnnn2 1n2Go back and read scenario (iii) of the LIMIT COMPARISON test carefully: because we got a LIMIT of , wecan only make a conclusion if we know that the corresponding series in the denominator diverges. But n=11n2. So the test is inconclusive!3 This warning shows that even though logarithmic growth is insanely slow, you cannot simply ignore do have to budget a small amount of growth to absorb the logarithm. This is where the heuristiclnn nsmall#comes into play. If we run this heuristic on our summand, we getlnnn2 nsmall#n2=1n2 (small#).Because we want to show that our series converges, we want to make sure that the series n=11n2 (small#)converges.

7 This means that we need2 (small#)>1and so0<small#<1. In this case, we can picksmall# =12and thus run LIMIT COMPARISON against1n32. Indeed,limn lnnn21n32= limn lnnn12= na lnnn32 1n32 Because we got a LIMIT of0and we know that n=11n32converges, we can successfully apply scenario (ii)of the LIMIT COMPARISON test to conclude that n=1lnnn2also Case III: Trig functions and other oddities (Taylor polynomials)The main take away from this case is the following approximation:sin(1n) does this come from, and are there other useful approximations?The first approximation:sin(1n) 1nLet s talk about the functionsinxfirst, and then I will describe the more general situation. Consider thegraph ofy= sinx. The tangent line atx= 0isy=x, becausesin(0) = 0andcos(0) = 1. We know that closeto0, this tangent line should be a good approximation to the function. Indeed, we can graph both near0toconfirm this: 2 112 2 112xyy= sinxy=x4It is clear thatsinx xas long asxis close to0.

8 The fraction1nis close to0for large values ofn, so indeedsin(1n) 1n. We can exploit this fact to generate useful candidates for LIMIT COMPARISON . For example,consider the series n=1sin(1n) on the above discussion, my intuition tells me that n=1sin(1n)n n=11nn= n= usual, the formalization of this intuition is LIMIT COMPARISON . Let s compute:limn sin(1n)n1n2= limn sin(1n)1n= limx sin(1x)1x(L H) = limx cos(1x) 1x2 1x2= cos(0) = a finite, positive number, we are in scenario (i) of the LIMIT COMPARISON test. This implies that theseries n=1sin(1n)nand n=11n2behave the same, just as we suspected. Since the latter is a convergentp-series, the former converges as well!Other useful approximationsCase III doesn t end with the approximation ofsin(1n). Becausesinx xforxsmall, we get a host of othersimilar approximations:sin(1n2) 1n2sin(1n3) 1n3sin(1 n) 1 nand so on and so forth. Furthermore, by manipulating these approximations we can go even further:sin2(1 n) (1 n)2=1n sin(1n3) 1n3= general, approximations likesinx xforxsmall can give you intuition about the behavior of manydifferent : The approximationsinx xonly holds whenxis small.

9 In order to use this approximation,the thing inside of the sine functionneeds to tend to0. In particular, if you see something likesin(n)in aseries,it would be wildly incorrect to claim thatsin(n) full story: Taylor polynomial approximationsMore generally, we can capture the behavior of other sequences by using knowledge of Taylor example, recall that the degree2 Taylor polynomial forcosxat0isT2(x) = 1 means that forxnear0, we have the approximationcosx 1 x22. We can use this approximation justlike we used the one forsinxabove. In particular, given a series like n=1(1 cos(1n2))we could identify the general behavior of the summand as1 cos(1n2) 12(1n2)2= could then use LIMIT COMPARISON to verify that n=1(1 cos(1n2))behaves the same as n=112n4, andthus another quick example of generating such approximations, recall that the degree1 Taylor polynomialforexat0isT1(x) = 1 + , we expectex 1 +xforxnear0. This leads to sequential approximations like1 e1n 1nand so on and so general description is this:given a series n=1f(sn)wherefis some function andsn 0, wecan generate a candidate for LIMIT COMPARISON by consideringTk(sn), whereTk(x)is a Taylor polynomialforfcentered Case IV: A last ditch resortIf you are given an infinite series and none of the above cases apply (and nothing else seems to work),sometimes you can just run LIMIT COMPARISON against won t always work, but manytimes it will.

10 Explicitly:- If you think the series n=1andiverges, run LIMIT COMPARISON If you think the series n=1anconverges, run LIMIT COMPARISON reason why this is an effective catch-all strategy is that in the world of convergent series, n=11n2isone of the "biggest" infinite series, , it converges pretty slowly. Likewise, n=11nis one of the smallestdivergent series, in that it diverges very slowly. Here s what I mean in a picture:6 Bigger sizeConvergent seriesDivergent series n=11n2 n=11nMost convergent seriesMost divergent seriesThis relative size COMPARISON is just a heuristic and not actually precise, but it is helpful nonetheless. Thepoint is this:if you want to conclude convergence via COMPARISON , you need to compare with some-thing bigger that also converges, and if you want to conclude divergence via COMPARISON , you need tocompare with something smaller that also diverges. Because n=11n2is heuristically bigger than mostconvergent series, it s a decent candidate for a convergent COMPARISON argument.