Convergence of series with ^(n+1) 
I want to check, if this series is convergent or not: 
$$\sum\limits_{n=1}^\infty  \frac{n^n(n!)}{(2n)!}$$ 
I tried with the ratio test:
$$\frac{\frac{(n+1)^{n+1}(n+1)!}{(2n+2)!}}{\frac{n^n*n!}{(2n)!} } =\frac{(n+1)^{n+1}(n+1)!(2n)!}{(2n+2)!(n!)n^n}=\frac{(n+1)^{n+1}(n+1)}{n^n(2n+2)(2n+1)}$$
But how should I proceed? 
Thank you for your help

queenD
 A: If you use Stirling approximation $$n! \approx \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ and applying it to the expression leads to $$\frac{n^n(n!)}{(2n)!} \approx 2^{-2 n-\frac{1}{2}} e^n=\frac{1}{\sqrt2} \Big(\frac{e}{4}\Big)^n$$ and then $$\sum\limits_{n=1}^\infty  \frac{n^n(n!)}{(2n)!} \approx  \frac{e}{\sqrt{2} (4-e)}$$
Added later
Computing the summation leads to a value of $1.53261$ and the approximation gives $1.49964$ which is not too bad.
A: Do you know the limit of $(1+1/n)^n$ and the limit of $(n+1)^2/(2n+2)(2n+1)$ as $n\to\infty$?
A: You are nearly finishing it:
$$\lim_{n\to \infty}\frac{(n+1)^{n+1}(n+1)}{n^n(2n+2)(2n+1)}=\frac{e}{4}<1$$
A: \begin{align}
\dfrac{n^n n!}{(2n)!} &= \dfrac{n^n}{(n+1)(n+2) \cdots 2n} \\
&= \prod_{k=1}^n \dfrac{n}{n+k} \leq \prod_{k= \lfloor n/2\rfloor + 1}^n \dfrac{n}{n+k} \\
&\leq \left(\dfrac{n}{n +  n/2}\right)^{n- \lfloor n/2\rfloor } = \left(\frac{2}{3}\right)^{n- \lfloor n/2\rfloor }\\ &\leq \left(\dfrac{2}{3}\right)^{n - n/2}=\left(\sqrt{\dfrac{2}{3}}\right)^n
\end{align}
A: I think this is the piece you're missing:
$$\begin{align}
\lim_{n \rightarrow \infty} \frac{(n+1)^n}{n^n} & = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n\\
\ & = e
\end{align}$$
