convergence of $\sum_{n=2}^{\infty}\frac{(n+1)!(n+1)^{n-1}}{n^{2n}}$ $$\sum_{n=2}^{\infty}\frac{(n+1)!(n+1)^{n-1}}{n^{2n}}$$
I used the Cauchy test and it lead me to $\frac{\sqrt[n]{n!}}{n^2}$. But I can't tell what is the limit of this. I tried the Squeeze theorem: 
$$\frac{\sqrt[n]{n}}{n^2} < \frac{\sqrt[n]{n!}}{n^2} < ??? $$
And I'm stuck.
 A: You could use AM-GM inequality. You get:
$$\sqrt[n]{n!} = \sqrt{1\cdot2\cdots n} \le \frac{1+2+\dots+n}n=\frac{n+1}2$$
which means
$$0\le \frac{\sqrt[n]{n!}}{n^2} \le \frac{n+1}{2n^2}$$ 
and thus
$$\lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n^2}=0.$$
A: You could just use the straight-forward ratio test:
$$ a_{n+1}/a_n = \frac{(n+2)! (n+2)^n n^{2n}}{(n+1)! (n+1)^{n-1} (n+1)^{2n+2}} = \left(\frac{n+2}{n+1}\right)^{n+1} \left(\frac{n}{n+1}\right)^{2n} $$
Since $\left(\frac{n+2}{n+1}\right)^{n+1} = \left(1+\frac{1}{n+1}\right)^{n+1} \to \mathrm{e}$ and $\left(\frac{n}{n+1}\right)^n = \left(1-\frac{1}{n+1}\right)^n \to \mathrm{e}^{-1}$, we have
$$a_{n+1}/a_n \to (\mathrm{e}) \left(\mathrm{e}^{-1}\right)^2 = \mathrm{e}^{-1} < 1. $$
A: Using the root test,
$\displaystyle\lim_{n\to\infty}\frac{\left((n+1)!\right)^{1/n}\left(n+1\right)^{1-1/n}}{n^2}=\lim_{n\to\infty}\frac{\left((n+1)!\right)^{1/n}}{n}\cdot\frac{n+1}{n}\cdot\frac{1}{(n+1)^{1/n}}=\lim_{n\to\infty}\frac{\left((n+1)!\right)^{1/n}}{n}$
$\hspace{2.2 in}$since $\displaystyle\frac{n+1}{n}\to1$ and $\displaystyle (n+1)^{1/n}\to 1$.
If $c_n=\displaystyle\frac{(n+1)!}{n^n}$, then  $\displaystyle\frac{c_{n+1}}{c_n}=\frac{(n+2)!}{(n+1)^{n+1}}\cdot\frac{n^n}{(n+1)!}=\frac{n+2}{n+1}\left(\frac{n}{n+1}\right)^n=\frac{n+2}{n+1}\frac{1}{(1+\frac{1}{n})^n}\to\frac{1}{e}$,
so $\displaystyle\lim_{n\to\infty}\frac{((n+1)!)^{1/n}}{n}=\lim_{n\to\infty}(c_n)^{1/n}=\frac{1}{e}<1$ and therefore the series converges.
