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Determine whether the series $\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}$ is convergent or divergent.

I've tried applying all the basic tests to no avail. I need to find out what the "trick" for this particular series is.

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    @ Ittay Weiss: the Divergence Test fails since the expression tends to $0$.2012-11-09

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Hint: Try with Cauchy's root test.

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    Ah, lesson learned. Thanks all!2012-11-09
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Let $u_n$ = $ \Big ( \frac {n}{n+1} \Big)^{n^2}$ , then since $ \ \lim_{n \to \infty}\sqrt[n]{({\frac{n}{n+1}})^{n^2}} = \lim_{n \to \infty}(\frac{n}{n+1})^n = \lim_{n \to \infty}(\frac{1}{1+\frac{1}{n}})^n$ = $\frac1e$ $ < 1 $ , so by Cauchy's root test the series is convergent .

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Note the following: $\Big(\frac{n}{n+1}\Big)^{n}=\Big(\frac{m-1}{m}\Big)^{m-1}=(1-1/m)^{m-1}$ That formula/function seems familiar, doesn't it?

Thereafter, note that: $\Big(\frac{n}{n+1}\Big)^{n^{2}}={\Big(\Big(\frac{n}{n+1}\Big)^{n}\space\Big)}^{n}$ Hint: Simple comparison test with a sequence you know well.

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    Nah, you could have used the root method. I just feel this is a lot more elegant.2012-11-09