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

  • $\begingroup$ How would you define x! ? Do you mean $\Gamma(x)$? $\endgroup$
    – Vim
    Sep 13 '15 at 11:02
  • $\begingroup$ x! = x(x-1)(x-2)...(3)(2)(1) $\endgroup$
    – bard
    Sep 13 '15 at 11:05
  • 2
    $\begingroup$ it is invalid when x is not an integer. $\endgroup$
    – Vim
    Sep 13 '15 at 11:06
  • $\begingroup$ But how would you integrate that ?. I mean x! $\endgroup$
    – user210387
    Sep 13 '15 at 11:07
  • $\begingroup$ '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 $\endgroup$
    – 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, $\Gamma(\frac{3}{2})$.

This, I repeat, is a pretty rough and not very elegant way of solving the problem. There are better ones too.


$\lim\limits_{n \to \infty}\sqrt[n]{\dfrac{n!}{n^n}}‎\sim \lim\limits_{n \to \infty} \dfrac{\dfrac{n}{e}}{n}=\dfrac{1}{e}<1$

  • $\begingroup$ Why is the nth root of n! n/e? $\endgroup$
    – bard
    Sep 13 '15 at 13:28
  • $\begingroup$ @bard, see "Stirling's approximation" $\endgroup$
    – user35603
    Sep 13 '15 at 13:48

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