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.
n=1n!nnn2
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Expert Answer

Paul Mitchell
Answered 2022-01-03 Author has 40 answers
Since by the Stirling approximation
nlnn!n2lnn(nlnn+o(n))n2lnn=n2(1o(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
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!)1n1n×n(n+1)2=n+122n3
for n large enough.
The term of your series is then to be compared to (23)n2

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