Evaluating $\lim_{n\to\infty} \left(\frac{n!}{n^n}\right)^{1/n}$ using Stirling's approximation

Question

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.

@TitoEliatron Could you cite some resource so that I can read about this proposition and where it comes from. — Paras Khosla, Nov 28 '20 at 20:34
How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$? has some answers that might be useful here. — robjohn, Nov 28 '20 at 21:23

score 1 · Accepted Answer · answered Nov 28 '20 at 20:33

1

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

answered Nov 28 '20 at 20:33

Tito Eliatron

6,612

Thanks for answering. Could you tell what this result is called? – Paras Khosla Nov 28 '20 at 20:39
I don0t know... is a classical result on limits of sequences. Here you can found a proof – Tito Eliatron Nov 28 '20 at 20:58
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This is essentially Stolz-Cesaro applied to the logarithm. – robjohn Nov 28 '20 at 21:25

Evaluating $\lim_{n\to\infty} \left(\frac{n!}{n^n}\right)^{1/n}$ using Stirling's approximation

1 Answers1