Juan Hewlett

2022-01-02

Show that the series converges.

$\sum _{n=1}^{\mathrm{\infty}}\frac{n{!}^{n}}{{n}^{{n}^{2}}}$

Paul Mitchell

Beginner2022-01-03Added 40 answers

Since by the Stirling approximation

$n\mathrm{ln}n!-{n}^{2}\mathrm{ln}\in n(n\mathrm{ln}-n+o\left(n\right))-{n}^{2}\mathrm{ln}n=-{n}^{2}(1-o\left(1\right))$ ,,

we have a superexponentially decaying upper bound on the n-th term, so the series converges.

we have a superexponentially decaying upper bound on the n-th term, so the series converges.

alkaholikd9

Beginner2022-01-04Added 37 answers

By Stirlings

karton

Expert2022-01-09Added 613 answers

If you don't want to use Stirling the simpler is to go for AM-GM inequality followed by rough majoration

for n large enough.

The term of your series is then to be compared to

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