4¢[ZBÝ YkBhc.'The geometric series and the ratio testkBhc«¤Today we are going to develop another test for convergence based on the interplay between the limit comparison test we developed last time and the geometric series.kBhc(!A note about the geometric serieskBhc›”Before we get into today's primary topic, I have to clear up a little detail about the geometric series. Here is a formula for the geometric series. muS(kBhc°kBhc n = 0kBhc a ropxEkBhcnkBhc  = carF(kBhcakBhc 1 - rkBhcD=An important detail to note here is that the sum starts with kBhc n = 0kBhc·°. Many times in what follows we will find ourselves having to look at variants of the geometric series that start at an index other than 0. These cases are very easy to handle. muS(kBhc°kBhc n = NkBhc a ropxEkBhcnkBhc  = muS(kBhc°kBhc n = NkBhc a ropxEkBhc n-NkBhc  ropxEkBhcNkBhc  = muS(kBhc°kBhc k = 0kBhc  a ropxEkBhckkBhc  ropxEkBhcNkBhc  = ropxEkBhcNkBhc muS(kBhc°kBhc k = 0kBhc  a ropxEkBhckkBhc  = ropxEkBhcNkBhc carF(kBhcakBhc 1 - r¢kBhc„}The one special trick needed here was a change in index. In going from the second series to the third, we replaced the index kBhcnkBhc with a new index kBhckkBhc where kBhc k = n - NkBhc. kBhcThe ratio test kBhcÿBoth the comparison test and the limit comparison test have a major drawback; namely, to use those tests to understand some series we have to be able to compare that series to something we already understand. In many cases this is difficult to do. PerhapskBhc/( the prime example of this is the series  muS(kBhc°kBhc n = 1carF=kBhc2opxEkBhcnkBhc n! kBhcþ÷The presence of the factorial term makes it difficult to compare this series with any other know series (or even to do an integral comparison). What we need to handle this case is some notion of convergence that exist independently of comparisons. kBhcRethinking convergencekBhcMFWhat does a convergent series look like? For starters, we know that if muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhcpiis to have any chance of converging, its terms must get small. However, as we have seen, merely requiringtmiLkBhc n Ï °kBhc  asbuSkBhcnkBhc  = 0kBhcƒ|is not sufficient. There are many examples of divergent series that do this, so the requirement that the terms get small as kBhcnkBhc% gets large is not conclusive.kBhcÿFor convergence to happen, the first requirement is that terms in the series have to get small. What we need is a slightly more clever way to express the idea that the terms are getting small. Here is that clever way to characterize the fact that the termkBhcs get smaller.tmiLkBhc nÏ°kBhc VsbAdcarFXkBhcasbuSkBhc n + 1kBhcasbuSkBhcnkBhc = L < 1kBhc‡€You have to stare for a while to convince yourself that this expresses the fact that the terms in series are getting smaller as kBhcnkBhc gets larger.kBhc)"Lets take as given that the limit feR (1)kBhc+$ exists and is less than 1. Pick an kBhcékBhc with the property thatkBhc L + é < 1kBhcunIf we go far enough out in the sequence of terms, the thing we are taking the limit of will stay smaller than kBhc L + ékBhc.kBhc For kBhc n > NkBhc we have VsbAdcarFXkBhcasbuSkBhc n + 1kBhcasbuSkBhcnkBhc < L + é < 1kBhcWe can rewritekBhc VsbAdcarFXkBhcasbuSkBhc n + 1kBhcasbuSkBhcnkBhc < L + ékBhc askBhcasbuSkBhc n + 1kBhc  < GrapkBhc L + ékBhc  asbuSkBhcnkBhc†Note that this is only legal if all the terms in the series were positive to begin with. Here is how we can use the inequality feR (2)kBhc  to get something useful.kBhcasbuSkBhc n + 1kBhc  < GrapkBhc L + ékBhc  asbuSkBhcn kBhcasbuSkBhc n + 2kBhc  < GrapkBhc L + ékBhc  asbuSkBhc n + 1kBhc  < GrapkBhc L + ékBhc GrapkBhc L + ékBhc  asbuSkBhcn!kBhc!More generally we get that"kBhcasbuSkBhc n + kkBhc  < GrapkBhc L + éopxEkBhckkBhc  asbuSkBhcn#kBhc0)This all allows you to say the following:$ muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhc  = muS(kBhcNkBhc n = 1kBhcasbuSkBhcnkBhc  + muS,kBhc°kBhc n = N + 1kBhcasbuSkBhcn% kBhc < muS(kBhcNkBhc n = 1kBhcasbuSkBhcnkBhc  + muS,kBhc°kBhc n = N + 1GrapkBhc L + éopxE3kBhc n - GrapkBhc N + 1kBhc  asbuSkBhc N+1& kBhc = muS(kBhcNkBhc n = 1kBhcasbuSkBhcnkBhc  +asbuSkBhc N+1kBhc muS,kBhc°kBhc n = N + 1GrapkBhc L + éopxE3kBhc n - GrapkBhc N + 1kBhc 'kBhcmfThe last sum that shows up here is the geometric series, and it shows that this whole thing converges.(kBhc4-The reasoning above leads us to the following)kBhcTheoremkBhc  Let muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhcC< be a series with only positive terms with the property that*tmiLkBhc n Ï °kBhc VsbAdcarFXkBhcasbuSkBhc n + 1kBhcasbuSkBhcnkBhc = L < 1+kBhc%Then the series must converge.,-kBhcMFThere is a closely related proof forthe divergent case, which leads to.kBhcTheoremkBhc  Let muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhcC< be a series with only positive terms with the property that/tmiLkBhc n Ï °kBhc VsbAdcarFXkBhcasbuSkBhc n + 1kBhcasbuSkBhcnkBhc = L > 10kBhc$Then the series must diverge.1kBhcA word of warning2kBhc ™The ratio test is very handy, and easy to apply, but there is one case in which it does not provide useful information. The test says that when the limit3tmiLkBhc n Ï °kBhc VsbAdcarFXkBhcasbuSkBhc n + 1kBhcasbuSkBhcn4kBhc¦Ÿis less than 1 the series will converge, and when the limit is greater than 1 the series will diverge. If the limit is exactly one, this test tells us nothing.5kBhcKDHere is an example to demonstrate this. Consider the harmonic series6 muS(kBhc°kBhc n = 1kBhc carF$kBhc1kBhcn7kBhcJCWe know that this series diverges. Applying the ratio test gives us8tmiLkBhc nÏ°carFAkBhc 1/GrapkBhc n+1kBhc 1/nkBhc  = tmiLkBhc nÏ°kBhc carF(kBhcnkBhc n + 1kBhc  = 19kBhc%Now consider convergent series: muS(kBhc°kBhc n = 1kBhc carF<kBhc1kBhcnopxEkBhc2;kBhc&In this case the ratio limit is<tmiLkBhc nÏ°carFqkBhc 1/GrapkBhc n+1opxEkBhc2kBhc 1/nopxEkBhc2kBhc  = tmiLkBhc nÏ°kBhc carFdkBhcnopxEkBhc2GrapkBhc n + 1opxEkBhc2kBhc  = 1=kBhc¢›These two examples demonstrate that the when the ratio limit is one the ratio test is completely inconclusive: the series could either converge or diverge.>kBhcExample?kBhcˆThe ratio test now allows to do some examples that would have been very difficult before. The best such example is the following.@ muS(kBhc°kBhc n = 1carF=kBhc2opxEkBhcnkBhc n!AtmiLkBhc nÏ°kBhc carF–kBhc2opxEkBhc n+1kBhc/GrapkBhc n+1kBhc!kBhc2opxEkBhcnkBhc /n!kBhc  = tmiLkBhc nÏ°kBhc carF–kBhc2opxEkBhc n+1kBhc  n!kBhc2opxEkBhcnkBhc GrapkBhc n+1kBhc!kBhc  =tmiLkBhc nÏ°kBhc  2 carFAkBhc n!GrapkBhc n+1kBhc  n!kBhc  = tmiLkBhc nÏ°carF(kBhc2kBhc n + 1kBhc  = 0BkBhc,%This shows that the series converges.CkBhcÇÀThis example demonstrates that the ratio test is especially well suited to handle series containing exponentials or factorials, as those things simplify particularly well when we take a ratio.DkBhcHere is a second exampleE muS(kBhc°kBhc n = 1carF<kBhcnkBhc2opxEkBhcnF tmiLkBhc nÏ°kBhc carFsGrapkBhc n+1kBhc /2opxEkBhc n+1kBhc n/2opxEkBhcnkBhc  = tmiLkBhc nÏ°kBhc carF‰kBhc2opxEkBhcnkBhc GrapkBhc n+1kBhc2opxEkBhc n+1kBhc  nkBhc  = tmiLkBhc nÏ°kBhc carF(kBhc n+1kBhc 2 nkBhc  = carF$kBhc1kBhc2GkBhc"This series also converges.HkBhcThe nth-root testIkBhc­¦As powerful and useful as the ratio test is, there are certain examples that are so nasty that something even stronger is required to solve them. Consider the exampleJ muS(kBhc°kBhc n = 1carFTkBhc3opxEkBhcnkBhcnopxEkBhcnKkBhc@9Applying the ratio test means having to compute the limitLtmiLkBhc nÏ°kBhc carF½kBhc3opxEkBhc n+1kBhc  /GrapkBhc n+1opxEkBhc n+1kBhc3opxEkBhcnkBhc / nopxEkBhcnkBhc  = ?MkBhc81This is not a particularly easy limit to compute.NkBhcýöA powerful test which often helps in cases like this one is the nth root test. I am not going to show the proof of the validity of the nth root test, but the methods of proof are very similar to those I used above to prove part of the ratio test.OkBhcTheorem kBhc(nth root test) Let muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhcC< be a series with only positive terms with the property thatPtmiLkBhc nÏ°kBhc trqS<kBhcnkBhcasbuSkBhcnkBhc  < 1QkBhc!Then the series converges.RkBhcTheorem kBhc(nth root test) Let muS(kBhc°kBhc n = 1kBhc  asbuSkBhcnkBhcC< be a series with only positive terms with the property thatStmiLkBhc nÏ°kBhc trqS<kBhcnkBhcasbuSkBhcnkBhc  > 1TkBhc Then the series diverges.UVkBhc?8When applied to the example at the start of this sectionW muS(kBhc°kBhc n = 1carFTkBhc3opxEkBhcnkBhcnopxEkBhcnXkBhcLEthe nth root test quickly and easily shows that the series converges.YtmiLkBhc nÏ°kBhc trqSpkBhcncarFTkBhc3opxEkBhcnkBhcnopxEkBhcnkBhc  = tmiLkBhc nÏ°kBhc carF$kBhc3kBhcnkBhc  = 0