Unformatted text preview: begins to look as if each term is 1 / 5 of the previous term. We have seen series that behave like this: ∞ X n =0 1 5 n = 5 4 , a geometric series. So we might try comparing the given series to some variation of this geometric series. This is possible, but a bit messy. We can in eﬀect do the same thing, but bypass most of the unpleasant work. The key is to notice that lim n →∞ a n +1 a n = lim n →∞ ( n + 1) 5 5 n +1 5 n n 5 = lim n →∞ ( n + 1) 5 n 5 1 5 = 1 · 1 5 = 1 5 . This is really just what we noticed above, done a bit more oﬃcially: in the long run, each term is one ﬁfth of the previous term. Now pick some number between 1 / 5 and 1, say 1 / 2....
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 Spring '07
 JonathanRogawski
 Math, Calculus, Sequences And Series, lim, one fifth

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