Juan Hewlett
2022-01-02
Answered

Show that the series converges.

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

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Paul Mitchell

Answered 2022-01-03
Author has **40** answers

Since by the Stirling approximation

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

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

Answered 2022-01-04
Author has **37** answers

By Stirlings

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

for n large enough.

The term of your series is then to be compared to

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Representing functions by power series Identify the functions represented by the following power series.

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If $\frac{1}{200}\sum _{n=1}^{399}\frac{{5}^{200}}{{5}^{n}+{5}^{200}}=\frac{a}{b}$ , then find $|a-b|$

asked 2022-01-21

Is it possible to write this in closed form:

$\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\mathrm{log}\left(\left(\begin{array}{c}n\\ k\end{array}\right)\right)$

asked 2022-05-02

Proved that question:

$S\left(n\right)=\sum _{k\ge 1}\frac{n!}{k(n-k)!{n}^{k}}=\sum _{k\ge 1}\frac{{n}^{\underset{\u2015}{k}}}{k{n}^{k}}\approx \frac{1}{2}\mathrm{log}\left(n\right)$

asked 2022-01-24

How to prove the following equality?

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{\prod _{k=1}^{m}(n+k)}=\frac{1}{(m-1)m!}$

asked 2022-01-18

Find the sum:

$S=\sum _{k=1}^{\mathrm{\infty}}\frac{1}{{k}^{2}-{a}^{2}}$

where$a\in (0,1)$

where

asked 2021-02-21

Use the Root Test to determine the convergence or divergence of the series.

$\sum _{n=1}^{\mathrm{\infty}}(\frac{3n+2}{n+3}{)}^{n}$