Finding Radius of convergence of $\sum_{n=2}^{\infty} \frac{n^{2n}}{4^n(2n+1)!} (3-2x)^n$ Radius of convergence of 
$$\sum_{n=2}^{\infty} \frac{n^{2n}}{4^n(2n+1)!} (3-2x)^n$$
The answer was to let $$c_n = \frac{(n^{2}(-2))^n}{4^n(2n+1)!}$$
But how do I get the $-2$? Its probably from $(3-2x)^n$. But how do I remove the $3$ and  $x$ parts? 
Also later into the answer, $$(\lim_{n\to\infty} ... (\lim_{n\to\infty} (1+\frac{1}{n})^n)^2) = \frac{e^2}{8}$$ 
What gives $e$ btw? 

I've seen something similar around. Since you need the $x$ part to be of the form $(x-a)^n$, you have to extract the $2$ in the $x$:
$$(3-2x)^n=(-2)^n \cdot (x-3/2)^n$$
Now you have
$$c_n = \frac{(-2n^2)^n}{4^n(2n+1)!}$$
$$c_n = \frac{(-2)^nn^{2n}}{4^n(2n+1)!}$$
You now need to evaluate the limit:
$$\lim\limits_{n \to \infty}  \left| \frac{(-2)^{n+1}{(n+1)}^{2(n+1)}}{(-2)^nn^{2n}}\frac{4^n(2n+1)!}{4^{n+1}(2n+3)!}\right|=$$
$$\lim\limits_{n \to \infty}\left| \frac{(-2){(n+1)}^{2n}{(n+1)}^2}{n^{2n}}\frac{1}{4(2n+2)(2n+3)}\right|=$$
$$\frac{1}{2}\lim\limits_{n \to \infty} \left|{\left(\frac{{n+1}}{n}\right)^n}^2\frac{n^2+2n+1}{4n^2+10n+6}\right|=$$
$$\frac{1}{2}\lim\limits_{n \to \infty}\left| {\left(1+\frac{{1}}{n}\right)^n}^2\frac{n^2+2n+1}{4n^2+10n+6}\right|=$$
This limits shouldn't be hard for you, I guess.

 $$\displaystyle \frac{1}{2}e^2\frac{1}{4}=\frac{e^2}{8}$$

