# Show that the series converges. \sum_{n=1}^\infty\frac{n!^n}{n^{n^2}}

Show that the series converges.
$\sum _{n=1}^{\mathrm{\infty }}\frac{n{!}^{n}}{{n}^{{n}^{2}}}$
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Paul Mitchell
Since by the Stirling approximation
$n\mathrm{ln}n!-{n}^{2}\mathrm{ln}\in n\left(n\mathrm{ln}-n+o\left(n\right)\right)-{n}^{2}\mathrm{ln}n=-{n}^{2}\left(1-o\left(1\right)\right)$,,
we have a superexponentially decaying upper bound on the n-th term, so the series converges.
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alkaholikd9
karton

If you don't want to use Stirling the simpler is to go for AM-GM inequality followed by rough majoration
$\left(n!{\right)}^{\frac{1}{n}}\le \frac{1}{n}×\frac{n\left(n+1\right)}{2}=\frac{n+1}{2}\le \frac{2n}{3}$
for n large enough.
The term of your series is then to be compared to $\left(\frac{2}{3}{\right)}^{{n}^{2}}$