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The n-th term of series is as follows

$$u_n = \left(\frac{n}{n+1}\right)^{n^2}$$

Does this series converge or diverge? **Since it was put on hold for "context". I am an engineer. I couldn't find a good way to attack this problem and thought would try friendly people at math exchange. Except that this question was put on hold I found it humbling to find so many elegant ways people came up to attack this problem from simple run of the mill test to solve it by ab initio principles.

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    $\begingroup$ yes the series does converge $\endgroup$ – Dr. Sonnhard Graubner Jul 10 '17 at 12:20
  • $\begingroup$ In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – Sahiba Arora Jul 10 '17 at 15:58
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Yes, it is convergent.

For each $n\in\mathbb N$, $\left(1+\frac1n\right)^n\geqslant2$. Therefore, $\left(1+\frac1n\right)^{n^2}=\left(\left(1+\frac1n\right)^n\right)^n\geqslant2^n$. So,$$\left(\frac n{n+1}\right)^{n^2}=\frac1{\left(1+\frac1n\right)^{n^2}}\leqslant\frac1{2^n}.$$

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  • $\begingroup$ Thanks! Very elegant from first principles. $\endgroup$ – manu datta Jul 10 '17 at 12:27
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$|u_n|^{1/n}=(\frac{n}{{n+1}})^{n}=(1-\frac{1}{n+1})^n$

Your turn !

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  • $\begingroup$ Cauchy's test ? $\endgroup$ – manu datta Jul 10 '17 at 12:21
  • $\begingroup$ No. better root test. $\endgroup$ – Fred Jul 10 '17 at 12:22
  • $\begingroup$ Root test. I think $\endgroup$ – Crazy Jul 10 '17 at 12:22
  • $\begingroup$ Thanks! Very nice solution. $\endgroup$ – manu datta Jul 10 '17 at 12:22
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Note that $e^x\geq1+x$, for $x\geq 0$

Then $u_n=\Big( 1+\frac{1}{n}\Big)^{-n^2}<(e^{\frac{1}{n}})^{-n^2}=e^{-n}$

Hence $\lim_{n\to\infty}u_n<\lim_{n\to\infty}e^{-n}=0\space\space\space\blacksquare$

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