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 $n \longrightarrow \infty$).
Best Answer
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 $x<e$.