121616949-math.272 - n →∞ | a n | 1/n = L If L< 1 the...

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258 Chapter 10 Sequences and Series This implies that | a N + k | > r k | a N | , but since r > 1 this means that lim k →∞ | a N + k | = 0, which means also that lim n →∞ a n = 0. By the divergence test, the series diverges. To see that we get no information when L = 1, we need to exhibit two series with L = 1, one that converges and one that diverges. It is easy to see that 1 /n 2 and 1 /n do the job. EXAMPLE 10.40 The ratio test is particularly useful for series involving the factorial function. Consider X n =0 5 n /n !. lim n →∞ 5 n +1 ( n + 1)! n ! 5 n = lim n →∞ 5 n +1 5 n n ! ( n + 1)! = lim n →∞ 5 1 ( n + 1) = 0 . Since 0 < 1, the series converges. A similar argument, which we will not do, justifies a similar test that is occasionally easier to apply. THEOREM 10.41 The Root Test Suppose that
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Unformatted text preview: n →∞ | a n | 1 /n = L . If L < 1 the series ∑ a n converges absolutely, if L > 1 the series diverges, and if L = 1 this test gives no information. The proof of the root test is actually easier than that of the ratio test, and is a good exercise. EXAMPLE 10.42 Analyze ∞ X n =0 5 n n n . The ratio test turns out to be a bit difficult on this series (try it). Using the root test: lim n →∞ ± 5 n n n ¶ 1 /n = lim n →∞ (5 n ) 1 /n ( n n ) 1 /n = lim n →∞ 5 n = 0 . Since 0 < 1, the series converges. The root test is frequently useful when n appears as an exponent in the general term of the series....
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