Hello everyone,

I tested the following series for convergence, but I'm not sure if all the steps I took are strictly legal. Thus I would like to ask you to take a look at them.

$\displaystyle \sum_{n=1}^{+\infty}{\frac{\left(n!\right)^2}{n^n} }$

I use the following fact to construct a series that is less than the one above for a comparison test.

$\displaystyle n!>\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$

$\displaystyle \sum_{n=1}^{+\infty} {\frac{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^2}{n^n}}$

Now I conduct a root test for this series to test it for convergence.

$\displaystyle \lim\limits_{n\rightarrow+\infty} {\frac{\sqrt[n]{2\pi n\left(\frac{n}{e}\right)^{2n}}}{\sqrt[n]{n^n}}}

= \lim\limits_{n\rightarrow+\infty} {\frac{\sqrt[n]{2\pi n}\left(\frac{n}{e}\right)^2}{n}}

= \lim\limits_{n\rightarrow+\infty} {\sqrt[n]{2\pi n}\frac{n}{e^2}}

= \lim\limits_{n\rightarrow+\infty} {e^{\frac{1}{n}\log{\left(2\pi n\right)}}\frac{n}{e^2}}

$

At this point the exponent of the exponential function is the problem, which I solve by using l'Hôpital's rule. (I'm not exactly sure if I am allowed to do this here.)

$\displaystyle \lim\limits_{n\rightarrow+\infty}{\frac{\log{\left (2\pi n\right)}}{n}} =

\lim\limits_{n\rightarrow+\infty} {\frac{2\pi}{2\pi n}}=\lim\limits_{n\rightarrow+\infty} {\frac{1}{n}}$

Now I move this newly obtained exponent back into the original limit calculation.

$\displaystyle \lim\limits_{n\rightarrow+\infty} {e^{\frac{1}{n}}\frac{n}{e^2}}=+\infty$

Since the smaller series diverges, also the greater one has to diverge.

I would appreciate it, if someone could tell me if I am right.