I tried using the Stirling's formula since $$ \frac{n^n}{n!}·\left(\frac1e\right)^n < \frac{n^{(n+0.5)}}{n!}·e^n $$ which equals to $(1/2\pi)$
I think this series converges but can't prove it...
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Try to use the Stirling formula the other way. Is any term of your series bigger than $\frac{1}{n}$ ? – Vincent Dec 19 '16 at 12:16
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Your insertion of Stirlings formula in your term seems faulty to a large degree. – Lutz Lehmann Dec 19 '16 at 13:31
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see also http://math.stackexchange.com/questions/2059754/ – user84413 Dec 19 '16 at 22:07
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You have $$u_n = \left( \frac{n}{e} \right)^n \frac{1}{n!} \sim \frac{1}{\sqrt{2 \pi n}}$$
according to Stirling formula which states $$n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$$
Hence the series diverges as $\sum \frac{1}{\sqrt{n}}$ diverges.
mathcounterexamples.net
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By Stirling's approximation, you have
$$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
Thus:
$$\frac{n^n}{n!e^n}\sim\frac{n^n e^n}{\sqrt{2\pi n}e^n n^n} = \frac{1}{\sqrt{2\pi n}}$$
The series you are considering is asymptotic to
$$\sum_{n=1}^{\infty}\frac{1}{\sqrt{2\pi n}} = \frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{n^{\frac{1}{2}}}$$ which is a $p$-series with $p = \frac{1}{2}\leq 1$ and is thus divergent. Therefore, your series diverges as well.
Tom
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