Convergence of Series w/ factorial I'm working on a convergence problem that's giving me trouble. I'll list the steps I've made so far. 
Given the following series determine if it is convergent or divergent:
$$\sum_{n=1}^{\infty}\frac{n!\cdot x^n}{n^n}, \text{where } x > 0.$$
When I first saw this problem I thought to use the root test so I attempted to preform the following calculation: 
$$\lim_{n\to\infty} \sqrt[n]{\left| \frac{n!\cdot x^n}{n^n} \right|}.$$
But here is where I'm not sure how to move forward. I'm basically unsure if we can distribute the $\frac{1}{n}$ exponent to $n!$ to generate something like this ${(n!)}^{1/n} \cdot x$ as the numerator (which would go to $x$ as $n \longrightarrow \infty$). 
 A: Here we can make use of Stirling's approximation, which gives us that $n!=\frac{n^n}{e^n}O(n)$, that is that $n!$ divided by $\frac{n^n}{e^n}$ is at most linear for large $n$. Thus $$\sum\limits_{n=1}^\infty \frac{n!\cdot x^n}{n^n}= \sum\limits_{n=1}^\infty \frac{n^n\cdot x^n\cdot O(n)}{e^n\cdot n^n}= \sum\limits_{n=1}^\infty \left(\frac{x}{e}\right)^nO(n)$$
which converges when $|x|<e$ by the ratio test. Using a more refined version of Stirling's, we get that $n!>c\frac{n^n}{e^n}$ for some constant $c>0$ and so 
$$\sum\limits_{n=1}^\infty \frac{n!\cdot x^n}{n^n}> \sum\limits_{n=1}^\infty \frac{n^n\cdot x^n\cdot c}{e^n\cdot n^n}= c\sum\limits_{n=1}^\infty \left(\frac{x}{e}\right)^n$$
which diverges when $|x|\geq e$. Thus the series converges if and only if $|x|<e$, so restricting our attention to $x>0$ we have that the series converges for $x<e$.
A: In lack of Stirlings approximation, one can use D'Alambert's criterion. We know a series converges if the ratio of its consecutive elements is eventually $<1$ in absolute value.
We have
$$ a_n=\frac{{n!}}{{{n^n}}}x^n$$
Then we find
$$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {x\frac{{{n^n}}}{{n!}}\frac{{\left( {n + 1} \right)!}}{{{{\left( {n + 1} \right)}^{n + 1}}}}} \right|$$
$$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^n}}}{{{{\left( {n + 1} \right)}^n}}}x} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{{\left( {\frac{n}{{n + 1}}} \right)}^n}x} \right|$$
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left| x \right|}}{{\left| {{{\left( {1 + \frac{1}{n}} \right)}^n}} \right|}} = \frac{{\left| x \right|}}{e}$$
Since we want the limit to be less than unity, it is manifest we need:
$$\frac{{\left| x \right|}}{e} < 1 \Rightarrow \left| x \right| < e$$
which is what Alex showed.
