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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1)^n * (2)^n * n!/n^N consider the seq of positive term and apply the ratio test

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can use the ratio test here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} (2)^n \frac{n!}{n^n}\] applying the ratio test theorem , we'll get: \[\frac{an+1}{an} = \frac{(2)^{n+1}(n+1)!}{(n+1)^{n+1}}\]\[ \.\frac{n^n}{n!.(2)^n}\]= \[=\frac{2^n.2^1.n!(n+1)}{(n+1)^n(n+1)}[\frac{n^n}{(2)^n .n!}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{2n^n}{(n+1)^n}\] 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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