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anonymous

  • 5 years ago

another fun series! (-2)^n * (n!/(n^n) ) as n goes from 1 to infinity...converging or diverging??..i cant use the ratio test rite?

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  1. anonymous
    • 5 years ago
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    (-1)^n * (2)^n * n!/n^N consider the seq of positive term and apply the ratio test

  2. anonymous
    • 5 years ago
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    you can use the ratio test here

  3. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty} (-2)^n \frac{n!}{n^n}\] applying the ratio test theorem , we'll get: \[|\frac{an+1}{an}| = |\frac{(-2)^{n+1}(n+1)!}{(n+1)^{n+1}}|\]\[ \.\frac{n^n}{n!.(-2)^n}\]= \[=\frac{-2^n.2^1.n!(n+1)}{(n+1)^n(n+1)}[\frac{n^n}{(-2)^n .n!}\]

  4. anonymous
    • 5 years ago
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    \[=\frac{2n^n}{(n+1)^n}\] after that find the limit and you'll compare your answers If L < 1 = convergent If L > 1 = divergent if L = 1, no conclusion ^_^ L = limit, you can take it from here :)

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