Ratio Test $\sum\limits_{n=0}^\infty\frac{(2n)^{2n}}{(2n)!}$ Im getting a little stuck on the following Ratio Test example... I haven't done it in so long, but it forms part of a question in a complex analysis past exam question
$\displaystyle\sum_{n=0}^\infty \dfrac{(2n)^{2n}}{(2n)!}$
The workings I have so far are.....
$\displaystyle\lim _{n \rightarrow \infty} \bigg|\dfrac{(2(n+1))^{2(n+1)}}{(2(n+1))!}.\dfrac{(2n)!}{(2n)^{2n}} \bigg| $
$\displaystyle\lim_{n\rightarrow\infty}\bigg| \dfrac{(2n+2)^{2n+2}}{(2n+2)!} . \dfrac{(2n)!}{(2n)^{2n}}\bigg|   $
$\displaystyle\lim_{n\rightarrow\infty}\bigg|\dfrac{(2n+2)^{2n+2}}{(2n)^{2n}}. \dfrac{(2n)!}{(2n+2)!}\bigg|    $
$\displaystyle\lim_{n\rightarrow\infty}\bigg|\dfrac{(2n+2)^{2n+2}}{(2n)^{2n}}.\dfrac{1}{(2n+2)(2n+1)}\bigg|$
But from here I don't know what to do..Any help would be greatly appreciated
 A: Continuing from your last step:
$$\ldots=\left(1+\frac1n\right)^{2n}\frac{2n+2}{2n+1}\xrightarrow[n\to\infty]{}e^2\cdot 1=e^2$$
and thus the series...
Added:
$$\lim_{n\rightarrow\infty}\left|\dfrac{(2n+2)^{2n+2}}{(2n)^{2n}}.\dfrac{1}{(2n+2)(2n+1)}\right|=\frac{(2n+2)^{2n}(2n+2)^2}{(2n)^{2n}}\cdot\frac1{(2n+1)(2n+2)}=$$
$$=\left(\frac{2n+2}{2n}\right)^{2n}\cdot\frac{(2n+2)^2}{(2n+1)(2n+2)}=\left(1+\frac 1n\right)^{2n}\frac{2n+2}{2n+1}\;\;\ldots$$
A: The problem you seem to be having is that you need some way of combining $(2n+2)^{2n+2}$ and $(2n)^{2n}$. (Have you analyzed your work enough to realize this?)
There are two easy ways to combine exponentials: when they have the same base, or when they have the same exponent.
Both approaches will work here: I could, for example, write
$$ (2n+2)^{2n+2} = (2n)^{2n+2} \cdot \left( \frac{2n+2}{2n} \right)^{2n+2} $$
and continue from there. How did I come up with this? I first wrote down on the right hand side what I wanted to turn the exponential into, then I added in more to force both sides to be equal; in this case, the most obvious approach was to add in a factor that cancels the new thing and replaces it with the old thing.
I suggest you try, on your own, to take the other approach: change the exponentials so they have the same exponent.
It turns out that, either way, the rest of the problem will be roughly the same.
A: It is $ \frac{(2n)^{2n}}{(2n)!} \geq 1$ for all $n$, so the sum doesn't converge, no?
(As $\sum a_n$ converges $\Rightarrow \lim a_n = 0$.)
A: $(2n)^{2n}=2n\cdot2n\cdot2n\cdots $($2n$ times)$\ge1\cdot2\cdot3\dots\cdot 2n$
So,$\dfrac{(2n)^{2n }}{(2n)!}\ge1$ ,so  the sum diverges.
Edit:
The OP has commented that this is part of the question, find the radius of convergence of $\sum_0^{\infty}a_kz^k$.
We know that the radius is $\lim_{n\to\infty}\dfrac{1}{\sqrt[n]a_k}$. Use the fact that $\dfrac{\sqrt [n]{n!}}{n}\to\dfrac{1}{e}$ to find this limit.
