# ratio vs. root test for $\sum_{n=1}^\infty \frac{n^n}{(n!)^2}z^n$

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.

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 $= \frac{e}{n}$. – AD. Jan 4 '11 at 7:20

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. – ftiaronsem Jan 3 '11 at 14:52

I'd use Stirling here.

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wow, thanks that was fast and absolutly right. Yeah, this really works. Thank you very much. – ftiaronsem Jan 3 '11 at 14:35
there is just one question I still have about this. Stirlings formulae is just an approximation, but for $n \to \infty$ it becomas exact, right? So I am always allowed to do this substitution in case I am looking at $n \to \infty$? Or are there cases where I have to take care of something? – ftiaronsem Jan 3 '11 at 14:43
@Shai linked to a page explaining how and when Stirling is valid for looking at limits as $n \to\infty$. – non-expert Jan 3 '11 at 14:56
@ftiaronsem: The difference between $n!$ and Stirling's approximation is of lower order than $n!$, in other words, this difference divided by $n!$ tends to $0$ as $n$ goes to infinity. This information is sufficient to conclude e.g. the ratio test you have. – timur Jan 3 '11 at 17:58
ahh thank you very much. Excellent to know that, I really love asking on this site. – ftiaronsem Jan 3 '11 at 18:04