Examine convergence of $ \sum^{\infty}_{n=2}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}$ I need to examine convergence of the following sum:
$$\displaystyle \sum^{\infty}_{n=2}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}$$
I know that:
$$\displaystyle \lim_{n \to \infty}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n} = 0$$
And that all elements of sequence $a_n$ are positive. Therefore I can use Cauchy's root test:
$$\sqrt[n]{\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}}= \frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n}$$
$$\sqrt[n]{\frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n}}= \frac{\sqrt[n^2]{n!}(1+\frac{1}{2n})}{\sqrt[n]{n}}$$
But it gives me noting since I don't know what happens to $\sqrt[n^2]{n!}$ when $n \to \infty$. I tried also d'alembert's ratio test, but it gives me nothing. I think it may be LTC on limits.

Hint:
$$\lim_{n \to \infty} \frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n} = \lim_{n \to \infty}\frac{\sqrt[n]{n!}}{n} \lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{n}$$
To justify the split of limit, you should note that
$$\lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{n}$$
is a common limit (and I'm sure you can find it). For the first limit, refer to here.
