The geometric series and the ratio test


 Horatio Haynes
 2 years ago
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1 The geometric series ad the ratio test Today we are goig to develop aother test for covergece based o the iterplay betwee the it compariso test we developed last time ad the geometric series. A ote about the geometric series Before we get ito today's primary topic, I have to clear up a little detail about the geometric series. Here is a formula for the geometric series. = 0 a r = a 1  r A importat detail to ote here is that the sum starts with = 0. May times i what follows we will fid ourselves havig to look at variats of the geometric series that start at a idex other tha 0. These cases are very easy to hadle. = N a r = = N a r N r N = a r k r N = r N k = 0 a r k = r N k = 0 a 1  r The oe special trick eeded here was a chage i idex. I goig from the secod series to the third, we replaced the idex with ew idex k where k =  N. 1 The ratio test Both the compariso test ad theit comparisotest have a majordrawback; amely, to use those tests to uderstadsome series we have to be able to compare that series to somethig we already uderstad. I may cases this is difficult to do. Perhaps the prime example of this is the series The presece of the factorial term makes it difficult to compare this series with ay other kow series (or eve to do a itegral compariso). What we eed to hadle this case is some otio of covergece that exist idepedetly of comparisos. Rethikig covergece What does a coverget series look like? For starters, we kow that if! 1
2 is to have ay chace of covergig, its terms must get small. However, as we have see, merely requirig œ = 0 is ot sufficiet. There are may examples of diverget series that do this, so the requiremet that the terms get small as gets large is ot coclusive. For covergece to happe, the first requiremet is that terms i the series have to get small. What we eed is a slightly more clever way to express the idea that the terms are gettig small. Here is that clever way to characterize the fact that the terms get smaller. œ + 1 = L < 1 (1) You have to stare for a while to covice yourself that this expresses the fact that the terms i series are gettig smaller as gets larger. Lets take as give that the it (1)exists ad is less tha 1. Pick a È with the property that L + È < 1 If we go far eough out i the sequece of terms, the thig we are takig the it of will stay smaller tha L + È. We ca rewrite For > N we have + 1 < L + È < < L + È as + 1 < (L + È) () Note that this is oly legal if all the terms i the series were positive to begi with. Here is how we ca use the iequality () to get somethig useful.
3 + 1 < (L + È) More geerally we get that + < (L + È) + 1 < (L + È) (L + È) This all allows you to say the followig: N < + N + k < (L + È) k N = + = N + 1 = +a N +1 = N + 1 (L + È)  (N + 1 ) a N +1 = N + 1 (L + È)  (N + 1 ) The last sum that shows up here is the geometric series, ad it shows that this whole thig coverges. The reasoig above leads us to the followig Theorem Let be a series with oly positive terms with the property that œ + 1 = L < 1 The the series must coverge. There is a closely related proof forthe diverget case, which leads to Theorem Let be a series with oly positive terms with the property that œ + 1 = L > 1 The the series must diverge. A word of warig 3
4 The ratio test is very hady, ad easy to apply, but there is oe case i which it does ot provide useful iformatio. The test says that whe the it œ + 1 is less tha 1 the series will coverge, ad whe the it is greater tha 1 the series will diverge. If the it is exactly oe, this test tells us othig. Here is a example to demostrate this. Cosider the harmoic series 1 We kow that this series diverges. Applyig the ratio test gives us Now cosider coverget series I this case the ratio it is 1/(+1) = œ 1/ œ 1 1/(+1) = œ 1/ œ + 1 = 1 ( + 1) = 1 These two examples demostrate that the whe the ratio it is oe the ratio test is completely icoclusive: the series could either coverge or diverge. Example The ratio test ow allows to do some examples that would have bee very difficult before. The best such example is the followig. +1 /(+1)! œ = /! œ +1! (+1)!! = œ! (+1)! = œ + 1 = 0 4
5 This shows that the series coverges. This example demostrates that the ratio test is especially well suited to hadle series cotaiig expoetials or factorials, as those thigs simplify particularly well whe we take a ratio. Here is a secod example This series also coverges. The throot test (+1)/ +1 (+1) œ / = = +1 œ +1 œ As powerful ad useful as the ratio test is, there are certai examples that are so asty that somethig eve stroger is required to solve them. Cosider the example 3 Applyig the ratio test meas havig to compute the it 3 +1 /(+1) +1 œ 3 / =? This is ot a particularly easy it to compute. A powerful test which ofte helps i cases like this oe is the th root test. I am ot goig to show the proof of the validity of the th root test, but the methods of proof are very similar to those I used above to prove part of the ratio test. Theorem (th root test) Let property that be a series with oly positive terms with the < 1 œ The the series coverges. 5
6 Theorem (th root test) Let property that be a series with oly positive terms with the > 1 œ The the series diverges. Whe applied to the example at the start of this sectio 3 the th root test quickly ad easily shows that the series coverges. œ 3 = œ 3 = 0 6
4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
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