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

Juan Hewlett 2022-01-02 Answered

Show that the series converges.

\sum_{n = 1}^{\infty} \frac{n!^{n}}{n^{n^{2}}}

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Paul Mitchell

Answered 2022-01-03 Author has 40 answers

Since by the Stirling approximation

n \ln n! - n^{2} \ln \in n (n \ln - n + o (n)) - n^{2} \ln n = - n^{2} (1 - o (1))

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

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alkaholikd9

Answered 2022-01-04 Author has 37 answers

By Stirlings

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karton

Answered 2022-01-09 Author has 368 answers

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

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Calculus and Analysis

Post Secondary