Although it is a very simple question, I am not able to get similar results using Stirling's approximation as obtained using Integration. Here is what I have attempted.

$$L=\lim_{n\to\infty} \left(\frac{n!}{n^n}\right)^{1/n}$$

$$\begin{aligned}L&=\lim_{n\to\infty}\left(\frac{\sqrt{2\pi n}}{e^{n}}\right)^{1/n}\\&=\lim_{n\to\infty}\frac{1}{e}n^{1/2n}\end{aligned}$$

I am not able to reason out as to how this will evaluate to $1/e$. Any hints are appreciated. Thanks.


1 Answer 1


$\lim_n \sqrt[n]{a_n}=\lim_n \frac{a_{n+1}}{a_n}$ if the last limits exists. So $$\lim_n (\sqrt{2\pi n})^{1/n}=\lim_n \frac{\sqrt{2\pi (n+1)}}{\sqrt{2\pi n}}=\sqrt{\lim_n \frac{n+1}{n}}=1$$

  • $\begingroup$ Thanks for answering. Could you tell what this result is called? $\endgroup$ Nov 28, 2020 at 20:39
  • $\begingroup$ I don0t know... is a classical result on limits of sequences. Here you can found a proof $\endgroup$ Nov 28, 2020 at 20:58
  • 2
    $\begingroup$ This is essentially Stolz-Cesaro applied to the logarithm. $\endgroup$
    – robjohn
    Nov 28, 2020 at 21:25

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