Convergence Test (Ratio Test) for $\sum_n \frac{(2n)!}{n^{2n}}$ I am having trouble figuring out how to tackle this problem, where I have to determine if the following sum is convergent. 
$$\sum_n \dfrac{(2n)!}{n^{2n}} $$
I am assuming that you would need to apply the ratio test here, but have no idea how to continue after this:
$$\dfrac{(2n+2)!}{(n+1)^{2(n+1)}} \dfrac{n^{2n}}{(2n)!} $$
How would i go about canceling some terms out?
 A: Simplifying, you get$$(2n+2)(2n+1)\frac{n^{2n}}{(n+1)^{2n+2}}=\frac{(2n+2)(2n+1)}{(n+1)^2}\left(\frac n{n+1}\right)^{2n}\to\frac4{e^2}.$$
A: Lets start by stating what the ratio test is. 
The ratio test is  given as:
$$L=\lim_{n\rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|$$
Here $a_n$ is given as the terms of the sum, i.e:
$$\sum_n a_n$$
We can thus identify from your problem that:
$$a_n=\frac{(2n)!}{n^{2n}}$$
and (the $2$´s arrive because $n\rightarrow n+1$ gives $2n\rightarrow2n+2$),
$$a_{n+1}=\frac{(2n+2)!}{(n+1)^{2n+2}}$$
Thus giving us:
$$L=\lim_{n\rightarrow \infty}\left| \frac{\frac{(2n+2)!}{(n+1)^{2n+2}}}{\frac{(2n)!}{n^{2n}}} \right|$$
Which can be rewritten as:
$$L=\lim_{n\rightarrow \infty}\left| \frac{(2n+2)!n^{2n}}{(n+1)^{2n+2}(2n)!} \right|$$
We can note that we can rewrite this as:
$$L=\lim_{n\rightarrow \infty}\left| \frac{(2n+2)!}{(n+1)^{2}(2n)!}\frac{n^{2n}}{(n+1)^{2n}} \right|$$
Here we can identify by the following:
$$\frac{(2n+2)!}{(2n)!}=(2n+1)(2n+2)$$
Now by insertion:
$$L=\lim_{n\rightarrow \infty}\left| \frac{(2n+1)(2n+2)}{(n+1)^{2}}\frac{n^{2n}}{(n+1)^{2n}} \right| = \lim_{n\rightarrow \infty}\left| \frac{(2n+1)(2n+2)}{(n+1)^{2}}\frac{n^{2n}}{(n+1)^{2n}} \right|$$
By expanding out:
$$L = \lim_{n\rightarrow \infty}\left| \frac{4n^2 + 6n + 2}{n^2 + 2n +1}\frac{n^{2n}}{(n+1)^{2n}} \right|$$
We can see that the first fraction is just the limit of two polynomials, thus giving:
$$L = \lim_{n\rightarrow \infty}\left| 4\frac{n^{2n}}{(n+1)^{2n}} \right|$$
Now identifying the following limit:
$$e^x = \lim_{k\rightarrow \infty}\left( 1 + \frac{x}{k} \right)^k \rightarrow (e^{x})^{-1} = \lim_{k\rightarrow \infty} \left(\frac{k}{k+x} \right)^k \rightarrow e^{-1} = \lim_{k\rightarrow \infty} \left(\frac{k}{k+1} \right)^k $$
Now by insertion with $k\rightarrow n$:
$$L=\lim_{n\rightarrow \infty}\left| 4\left(\frac{1}{e^2}\right) \right| = \frac{4}{e^2}$$
A: One more way, if you use Stirling's approximation for $2n$: 
$$
(2n)! \sim \bigg(\frac{2n}{e} \Bigg)^{2n} 4 \sqrt{\pi n}
$$
then $n^{2n}$ term cancels out and you are left with (set $\frac{4}{e^2} = x <1$)
$$
4 \sqrt{\pi} \sum_{k=1}^{n} \bigg(\frac{4}{e^2}\bigg)^k \sqrt{k} < 4 \sqrt{\pi} \sum_{k=1}^{n}k x^k \to \frac{ 4 \sqrt{\pi} x}{(1-x)^2}
$$
this convergence works since $\frac{4}{e^2} <1$. Since the original sum is upper-bounded by the convergent sum, it converges too.
A: The root test is simpler.
Since
$(2n)! \approx \sqrt{cn}(2n/e)^{2n}$,
and $n^{1/n} \to 1$,
the n-th root goes to
$\dfrac{4n^2/e^2}{n^2}
= 4/e^2 < 1$
so the sum converges.
If the terms are
$(kn)!/n^{kn}
\approx \sqrt{ckn}(kn/e)^{kn}/n^{kn}
=\sqrt{ckn}(k/e)^{kn}$,
the n-th root goes to
$(k/e)^k$
so the sum converges for
$k<e$ and diverges for
$k>e$,
A: $$\frac{(2n+2)!}{(n+1)^{2(n+1)}}\frac{n^{2n}}{(2n)!}=\frac{(2n)!(2n+1)(2n+2)}{(n+1)^{2n}(n+1)^2}\frac{n^{2n}}{(2n)!}=\frac{2(2n+1)}{n+1}(\frac{n}{n+1})^{2n}$$
$$\lim_{n\to+\infty}\frac{2(2n+1)}{n+1}=4$$
$$\lim_{n\to+\infty}(\frac{n}{n+1})^{n}=\lim_{n\to+\infty}(\frac{n+1-1}{n+1})^{n}=\lim_{n\to+\infty}((1+\frac{-1}{n+1})^{-(n+1)})^{-\frac{n}{n+1}}=\lim_{n\to+\infty}e^{-\frac{n}{n+1}}=\frac{1}{e}$$
therefore
$$\frac{2(2n+1)}{n+1}(\frac{n}{n+1})^{2n} = \frac{4}{e^2} < 1$$
So the sum converges.
