Holetaug

2022-07-14

Why the limit of the following sequence is like this:
$\frac{2\sqrt[n]{n!}}{n}=\frac{2}{e}$

Jaelynn Cuevas

Expert

We can use the fact that if ${a}_{n}>0$ for all $n\ge 1$ and the sequence $\frac{{a}_{n+1}}{{a}_{n}}$ converges in $\left[0,\mathrm{\infty }\right]$, then
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}^{1/n}=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n+1}}{{a}_{n}}$
We have that ${a}_{n}=n!/{n}^{n}$ and
$\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n+1}}{{a}_{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\left(n+1\right)!{n}^{n}}{\left(n+1{\right)}^{n+1}n!}=\underset{n\to \mathrm{\infty }}{lim}\frac{{n}^{n}}{\left(n+1{\right)}^{n}}=\frac{1}{\underset{n\to \mathrm{\infty }}{lim}\left(1+1/n{\right)}^{n}}=\frac{1}{e}.$

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