Integral of $\frac{x!}{x^x}$

Question

I want to use the integral test to check if the following series converges (it does) and not sure how to integrate $f(x)$.

$$ \sum_{n=1}^\infty\frac{n!}{n^n} $$

How should I find $\int \frac{x!}{x^x}\,dx$?

Answer 1

By Stirling's approximation, $n! \sim \frac{n^n}{e^n} \sqrt{2 \pi n}$. Therefore, $n! / n^n \sim \frac{\sqrt{2 \pi n}}{e^n}$.

Therefore, for all $\epsilon > 0$, there is a $n_0$ such that for all $n > n_0, n! / n^n < \frac{\sqrt{2 \pi n}}{e^n}(1 + \epsilon)$.

Since $\sum_{n=1}^\infty \frac{\sqrt{2 \pi n}}{e^n}(1 + \epsilon)$ converges, so does $\sum_{n=1}^\infty \frac{n!}{n^n}$