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

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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did you try the divergence test? – Ittay Weiss Nov 9 '12 at 5:09
@JavaMan Does this hint help? That sequence goes to $0$ as $n$ tends to infinity. – Benjamin Dickman Nov 9 '12 at 5:11
@F'OlaYinka - Are you sure? It doesn't seem to me that your inequality holds for $n\geq{3}$. You want to say that $\frac{2}{3}\leq{\frac{1}{2}}$? – Johnny Westerling Nov 9 '12 at 5:17
@B.D. Thanks for the comment. I misread the sequence as $a_n = \left( \frac{n+1}{n}\right)^{n^2}$ for some reason. – JavaMan Nov 9 '12 at 5:36
@ Ittay Weiss: the Divergence Test fails since the expression tends to $0$. – The Substitute Nov 9 '12 at 6:24

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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Oh, snap - that was the "trick" I was looking for. Thanks, Much appreciated. – Seeker Nov 9 '12 at 6:22
Nah, you could have used the root method. I just feel this is a lot more elegant. – GregRos Nov 9 '12 at 6:22

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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Hint: Try with Cauchy's root test.

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$\displaystyle \left(1+\frac1n\right)^n$ does not approach $1$ as $n\to\infty$. For one thing, directly expanding it gives that $\displaystyle \left(1+\frac1n\right)^n\ge 1+n\left(\frac1n\right)=2$. – Andrés E. Caicedo Nov 9 '12 at 6:15
@ Seeker : The limit is $\frac 1e$ $<1$ , so Cauchy's root test in fact proves the covergence – Souvik Dey Nov 9 '12 at 6:17
Ah, lesson learned. Thanks all! – Seeker Nov 9 '12 at 6:24