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.
 A: 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.
A: An alternative method. First note that
$$
\left( {1 + \frac{1}{{2n}}} \right)^{n^2 }  = \left( {\left( {1 + \frac{1}{{2n}}} \right)^{2n} } \right)^{n/2}  < e^{n/2} 
$$
for all $n\geq 1$. Second,
\begin{align*}
\log n! & = \sum\limits_{k = 1}^n {\log k}  = \log n + \sum\limits_{k = 1}^{n - 1} {\log k}  < \log n + \sum\limits_{k = 1}^{n - 1} {\int_k^{k + 1} {\log tdt} } \\ & = \log n + \int_1^n {\log tdt}  = (n + 1)\log n - n + 1.
\end{align*}
Thus
$$
n! < en\frac{{n^n }}{{e^n }} \Rightarrow \frac{{n!}}{{n^n }}\left( {1 + \frac{1}{{2n}}} \right)^{n^2 }  < ene^{ - n/2} .
$$
But
$$
\sum\limits_{n = 1}^\infty  {ne^{ - n/2} }  = \frac{{\sqrt e }}{{(\sqrt e  - 1)^2 }} \approx 3.9
$$
converges, so does the original series.
A: Since $n!^{1/n} \approx n/e$, the n-th root is about
$(n/e)(1+1/(2n))^n/n
\approx e^{1/2}/e
= e^{-1/2}<1$
so the sum converges.
