# Prove that the series $\frac{x!}{x^x}$ converges using the integral test

The series $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ is clearly positive and decreasing, but how does one go about integrating $\int\frac{x!}{x^x} dx$ ?

• How would you define x! ? Do you mean $\Gamma(x)$?
– Vim
Sep 13 '15 at 11:02
• x! = x(x-1)(x-2)...(3)(2)(1)
– bard
Sep 13 '15 at 11:05
• it is invalid when x is not an integer.
– Vim
Sep 13 '15 at 11:06
• But how would you integrate that ?. I mean x!
– user210387
Sep 13 '15 at 11:07
• 'The series...is decreasing' That's not true: the series is increasing because every term is positive. What you mean is the sequence $\frac{k!}{k^k}$ is decreasing
– Alex
Sep 13 '15 at 12:49

A very rough way of showing convergence is use Stirling's expansion of the numerator. Since $k! \sim (\frac{k}{e})^k \sqrt{2 \pi k}$ the summand becomes $a_k \sim \frac{\sqrt{2 \pi k}}{e^k}$ which you can then compare to the integral, $\Gamma(\frac{3}{2})$.
$\lim\limits_{n \to \infty}\sqrt[n]{\dfrac{n!}{n^n}}‎\sim \lim\limits_{n \to \infty} \dfrac{\dfrac{n}{e}}{n}=\dfrac{1}{e}<1$