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I am trying to calculate the radius of convergence of

$\sum_{n=1}^\infty \frac{n^n}{(n!)^2}z^n$

with $z \in \mathbb{C}$. According to WolframAlpha, both the root and the ratio test are conclusive, indicating convergence. However at the root test, I am stuck at:

(For convenience I drop the limes superior in the following equations, but it should be in front of every expression.)

$\sqrt[n]{\frac{n^n}{(n!)^2}}=\frac{n}{\sqrt[n]{n!}\sqrt[n]{n!}}$

Is there some estimate of $\sqrt[n]{n!}$ I should know of? For the ratio test, I think I get a reasonable solution:

$\frac{\frac{(n+1)^{n+1}}{(n+1)!^2}}{\frac{n^n}{n!^2}}=\frac{(n+1)^{n+1}n!^2}{(n+1)!^2n^n} = \frac{(n+1)^{n-1}}{n^n} = \frac{1}{n}(1+\frac{1}{n})^n = \frac{e}{n}$

which results in a radius of convergence of $\infty$. I should get the same result with the root test, shouldn't I? After all the root test is stronger.

Hopefully someone can point out of how to continue with the root test.

Thanks in advance

ftiaronsem

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    There are two typos in the last expression: $\frac{1}{n+1}$ instead of $\frac{1}{n}$ and $\lt \frac{e}{n+1}$ instead of $= \frac{e}{n}$.2011-01-04

2 Answers 2

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You can use $n! > (\frac{n}{e})^n$. See this for an elementary proof of this result.

EDIT: Actually, this result can be easily proved using induction.

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    yeah, thank you very much. This also works great and I know when I am allowed to use it ;-). Thanks.2019-04-20
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I'd use Stirling here.

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    Don't forget to answer others' questions and earn points!2011-01-03