4D¢O(),¢¡ Ÿžœ›š™˜—–•”“’‘ŽŒ‹Š‰ˆ‡†…„ƒ‚€~}|l‹Šijmnopqrstuvwxy{|}~€‚ƒ„…†‡ˆŒŽeCùö dkBhcSeries ComparisonskBhc Simple series comparisonskBhcleThe next theorems demonstrate how to use known series to learn things about closely related examples.kBhcTheoremkBhc  If muS(kBhc°kBhc n = 0kBhcasbuSkBhcnkBhc converges and kBhc 0 ² bsbuSkBhcnkBhc  ² asbuSkBhcnkBhc for all kBhcnkBhc, then muS(kBhc°kBhc n = 0kBhcbsbuSkBhcnkBhc converges, too.kBhcTheoremkBhc  If muS(kBhc°kBhc n = 0kBhcasbuSkBhcnkBhc diverges and kBhcbsbuSkBhcnkBhc  ³ asbuSkBhcnkBhc  ³ 0kBhc for all kBhcnkBhc, then muS(kBhc°kBhc n = 0kBhcbsbuSkBhcnkBhc diverges, too.kBhcExamplekBhc Consider the series muS(kBhc°kBhc n = 1carF>kBhc1kBhc n 2opxEkBhcn kBhcg`We can understand this series by making a comparison with the known, convergent geometric series muS(kBhc°kBhc n = 1carF<kBhc1kBhc2opxEkBhcnkBhc  = muS(kBhc°kBhc n = 1kBhcropxEkBhcnkBhc  = carF(kBhc1kBhc 1 - rkBhc  = carF*kBhc1kBhc1 - 1/2kBhc  = 2 kBhcBecause kBhc 0 < carF>kBhc1kBhc n 2opxEkBhcnkBhc  < carF<kBhc1kBhc2opxEkBhcn kBhcfor all kBhcnkBhcZS, we see that the series converges, and we even get a crude estimate for its value.kBhc 0 < muS(kBhc°kBhc n = 1carF>kBhc1kBhc n 2opxEkBhcnkBhc  < 2kBhcš“In order to apply this method, we need to have a list of ÒknownÓ series to use as the basis for our comparisons. These are the things that we know: muS(kBhc°kBhc n = 1kBhcropxEkBhcnkBhc  = carF(kBhc1kBhc 1 - rkBhc  for VsbAkBhcrkBhc  < 1 muS(kBhc°kBhc n = 1kBhc carF$kBhc1kBhcnkBhc kBhcdiverges muS(kBhc°kBhc n = 1kBhc carF<kBhc1kBhcnopxEkBhc2kBhc converges muS(kBhc°kBhc n = 1kBhc carF<kBhc1kBhcnopxEkBhcpkBhc  = FlpsakBhc convergeskBhcdivergeskBhc p > 1kBhc p ² 1kBhc¡šIn other cases, you can use the integral comparison test to understand some ÒsimpleÓ series, and then use series comparison from there to tell you things.kBhc2+An example to motivate further developmentskBhcŒ…Next we are going to see an example which demonstrates that although the comparison test is powerful, it is often difficult to apply.kBhcConsider the series muS(kBhc°kBhc n = 1carFEkBhc n + 1GrapkBhc n + 2kBhc  nkBhcleThe first thing you should notice about this series is that it strongly resembles the harmonic series muS(kBhc°kBhc n = 1kBhc carF$kBhc1kBhcnkBhc­¦which we know to diverge. That part is easy. The difficulty comes when you notice that the most simple and obvious comparison you can write down is not at all useful. muS(kBhc°kBhc n = 1carFEkBhc n + 1GrapkBhc n + 2kBhc  nkBhc  < muS(kBhc°kBhc n = 1kBhc carF$kBhc1kBhcnkBhc  = °kBhcÿSaying that a series is less than a series which is known to diverge is not useful information. It tells us nothing about the series in question. The way out of this impasse is to concoct a trickier comparison which does tell us something useful. ConsiderkBhc the comparisoncarF$kBhc1kBhc2kBhc muS(kBhc°kBhc n = 1kBhc carF$kBhc1kBhcnkBhc  < muS(kBhc°kBhc n = 1carFEkBhc n + 1GrapkBhc n + 2kBhc  n kBhc6/The comparison is based on the observation that!carF$kBhc1kBhc2kBhc  < carF,kBhc n + 1kBhc n + 2"kBhcfor all kBhc n ³ 1kBhc~w. The comparison is a useful one, because 1/2 times the harmonic series diverges just as surely as the harmonic series.# kBhc ° = carF$kBhc1kBhc2kBhc muS(kBhc°kBhc n = 1kBhc carF$kBhc1kBhcnkBhc  < muS(kBhc°kBhc n = 1carFEkBhc n + 1GrapkBhc n + 2kBhc  n$kBhc5.Thus we see that the original series diverges.%kBhc‰‚This example demonstrates that although the comparison test is useful, it often requires inordinate cleverness to apply correctly.&kBhc+$A simpler method based on comparison'kBhcÿNext we are going to develop a technique based on the comparison method, but which does not require nearly as much cleverness. I am going to make an effort to manufacture this method from first principles. This is a valuable exercise, in that it should gikBhcvove you some valuable insight into the workings of this method and other methods which are coming down the line.*kBhc6/We begin the discussion with a vague but useful+kBhc Basic ideakBhc~w If a series Ôlooks likeÕ something that converges or diverges, the series must also converge or diverge, respectively.-kBhcµ®The most obvious question that this statement raises is ÒWhat does it mean for one series to Ôlook likeÕ another?Ó Here is one possible way to express this concretely. Saying. muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhc looks like muS(kBhc°kBhc n = 1kBhc  bsbuSkBhcn/kBhcbasically means0kBhcasbuSkBhcnkBhc looks like kBhcbsbuSkBhcn1kBhc or2VsbASkBhcbsbuSkBhcnkBhc  - asbuSkBhcnkBhc is about 03kBhc?8More concretely, you can state this in terms of a limit.4tmiLkBhc nÏ°kBhc VsbASkBhcbsbuSkBhcnkBhc  - asbuSkBhcnkBhc  = 05 kBhcCkBhcŒ…The supposed theorem would have us conclude that the second series converges. This is clearly not the case, so our claim is not true.?@kBhcŸ˜Can we salvage anything from this failure? Is there another way to express the notion that one series Ôlooks likeÕ another series? Here is one other wayAtmiLkBhc nÏ°kBhc VsbA`carFTkBhcasbuSkBhcnkBhcbsbuSkBhcnkBhc  = 1BkBhcslThe idea is that two series Ôlook likeÕ each other if the ratio of their terms approaches 1 as n gets large.C kBhcC NkBhc we haveVkBhck - ¶ < VsbA`carFTkBhcasbuSkBhcnkBhcbsbuSkBhcnkBhc < k + ¶WkBhcG@The following steps demonstrate how that inequality can help us.XkBhc muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhc  = muS(kBhcNkBhc n = 1kBhc  asbuSkBhcnkBhc  + muS,kBhc°kBhc n = N + 1kBhcasbuSkBhcnkBhc  = muS(kBhcNkBhc n = 1kBhc  asbuSkBhcnkBhc  + muS,kBhc°kBhc n = N + 1kBhcbsbuSkBhcnkBhc carFTkBhcasbuSkBhcnkBhcbsbuSkBhcnY kBhc < muS(kBhcNkBhc n = 1kBhc  asbuSkBhcnkBhc  + muS,kBhc°kBhc n = N + 1kBhcbsbuSkBhcnkBhc VsbA`carFTkBhcasbuSkBhcnkBhcbsbuSkBhcnZ kBhc < muS(kBhcNkBhc n = 1kBhc  asbuSkBhcnkBhc  + muS,kBhc°kBhc n = N + 1kBhcbsbuSkBhcnkBhc GrapkBhc k + ¶[ kBhc < muS(kBhcNkBhc n = 1kBhc  asbuSkBhcnkBhc  + GrapkBhc k + ¶kBhc  muS,kBhc°kBhc n = N + 1kBhcbsbuSkBhcnkBhc \kBhc Since the kBhcbsbuSkBhcnkBhcâÛ series converges, everything in the last line converges. We have thus shown that the original series converges, because we have trapped it underneath some terms which are known to be finite. Thus the theorem is proved.]^kBhcˆThere is a similar version of this theorem for divergent series. I leave it to you to look up the statement of that theorem in the text._`kBhc0)Let's bring back the original example nowa muS(kBhc°kBhc n = 1carFEkBhc n + 1GrapkBhc n + 2kBhc  nbkBhcCompare this to c muS(kBhc°kBhc n = 1carF$kBhc1kBhcndkBhc via the limitetmiLkBhc nÏ°kBhc VsbAocarFccarFEkBhc n + 1GrapkBhc n + 2kBhc  nkBhc 1/nkBhc  = 1fkBhcXQSince the limit is 1, the divergent version of the theorem shows that the series feR a(2)kBhcwp diverges. Note that we accomplished this without any of the cleverness needed to apply the raw comparison test.g