Find a radius of convergence of power series I have to Find a radius of convergence of this power series

I' ve decided to use D'alambert indication:

Looking for a limit i meet a problem with a factorial

Please. help me finish this example and find a radius of convergence. Thanks!
 A: Be careful. $|\frac{a_n}{a_{n+1}}|= \frac{x^nn^n4^{n+1}(n+1)!}{x^{n+1}(n+1)^{n+1}4^n n!}=\frac{4n^n}{x(n+1)^n} \to \frac{4}{ex}$.
A: Use the cauchy hadamard criterion;  http://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem
$$\lim _{n \to \infty} \sqrt[n] {\frac{n^n}{4^n n!}} = \lim_{n \to \infty} \frac{n}{4 \sqrt[n]{n!}}$$
Since $$n! \sim \big (\frac{n}{e} \big )^{\ n} \sqrt{2\pi n}$$
we have $$\lim_{n \to \infty} \sqrt[n]{n!} = \frac{n}{e}$$ and substituting in the limit above you find your radius to be
$$\rho = \frac{4}{e}$$
A: I don't think you have the right test, it should be:
$$
\lim_{n \rightarrow \infty} \left|\frac{a_{n + 1}}{a_n}\right| < 1
$$
Think about it: the terms should be getting smaller and smaller (to converge) to the next term should be smaller than the previous not the other way around (unless you were looking for the ratio to be $>1$).  Using properties of factorials and exponentials (namely $(n + 1)! = (n + 1)n!$ and $a^{n + 1} = aa^n$):
$$
\left|\frac{a_{n + 1}}{a_n}\right| = \frac{(n + 1)^{n + 1}}{4n^n(n + 1)}x 
$$
We need to find:
$$
\lim_{n\rightarrow \infty}\frac{(n + 1)^{n + 1}}{4n^n(n + 1)} = \lim_{n\rightarrow \infty}\frac{(n + 1)^{n}}{4n^n} = \frac{1}{4}\lim_{n\rightarrow \infty}\left(\frac{n + 1}{n}\right)^n
$$
You may recognize this as the constant $e$ but if not, we can use L'Hospital's rule but first we need to do some simplification:
$$
\left(\frac{n + 1}{n}\right)^n = e^{n\ln\left(\frac{n + 1}{n}\right)}
$$
Now take the limit of the exponent:
$$
\lim_{n\rightarrow \infty} \frac{\ln(n + 1) - \ln(n)}{\frac{1}{n}} = \lim_{n\rightarrow \infty} \frac{\frac{1}{n + 1} - \frac{1}{n}}{-\frac{1}{n^2}} \\
\lim_{n\rightarrow\infty} \frac{n - (n + 1)}{-\frac{1}{n^2}n(n + 1)} = \lim_{x\rightarrow\infty} \frac{n}{n + 1} = 1
$$
Therefore:
$$
\lim_{n\rightarrow\infty} \frac{(n + 1)^{n + 1}}{4n^n(n + 1)}x  = \frac{x}{4} e
$$
Therefore the series converges when $|x| < \frac{4}{e}$.
