Interval of convergence of series I am trying to find the interval of convergence of 
$$\sum_{n=0}^{\infty}\frac{(nx)^n}{n!}$$
I tried letting $a_n=\frac{(nx)^n}{n!}$ and considered:
$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}\frac{(x(n+1))^{n+1}n!}{(n+1)!(xn)^n}$$
$$=\lim_{n \to \infty}\frac{x}{n+1}\frac{(n+1)^{n+1}}{n^n}$$
$$=\lim_{n \to \infty}\frac{x}{n+1}\frac{1}{\frac{n^n}{(n+1)^{n+1}}}$$
Not really sure how to show it, but wouldn't 
$$\lim_{n \to \infty}\frac{n^n}{(n+1)^{n+1}}=1$$
and 
$$\lim_{n \to \infty}\frac{x}{n+1}=0$$
which means the interval of convergence is infinite? I am not too confidant on this reasoning though and some help would be appreciated 

First, you need to remember the absolute values when using the ratio test, so you should be computing the limit $\displaystyle \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$. Then your computation is good to this point:
$$\lim_{n \to \infty}\left|\frac{x}{n+1}\frac{(n+1)^{n+1}}{n^n}\right|$$
You can simplify this to 
$\begin{align} &=\lim_{n \to \infty}|x|\frac{(n+1)^{n}}{n^n}\\
&=\lim_{n \to \infty}|x|\left(\frac{n+1}{n}\right)^n\\
&=\lim_{n \to \infty}|x|\left(1+\frac{1}{n}\right)^n\\
&=e|x|\end{align}$
