How does the series expansion of $\frac{n}{\sqrt[n]{n!} }$ equal $e$?

Question

I am trying to figure out how the series expansion of $\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}\ \text{equals}\ e$?

Answer 1

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots$

From this expansion, by setting x=n,

$e^n=1+\frac{n}{1!}+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots$

It is obvious that

$e^n \geq \frac{n^n}{n!}$

Also, we can take log

$\ln{\frac{n^n}{n!}}=n \ln n - (\ln{1}+\ln{2}+\cdots+\ln{n}) =n \ln n-\sum_{k=1}^{n} \ln k$

We can find the inequality using integration of $\ln x$

$\int_{1}^{n + 1}\ln x \; dx= [x \ln x - x]_{1}^{n+1} = (n + 1) \ln (n + 1) - n \geq \sum_{k=1}^{n} \ln k$

which leads to

$\ln{\frac{n^n}{n!}} \geq n \ln n - (n+1) \ln (n+1) + n$

and

$\frac{n^n}{n!} \geq \frac{n e^{2n}}{(n+1)e^{n+1}} = \frac{n e^{n-1}}{(n+1)}$

Therefore,

$e^n \geq \frac{n^n}{n!} \geq \frac{n e^{n-1}}{(n+1)}$

Taking nth root

$e \geq \frac{n}{\sqrt[\leftroot{-2}\uproot{2}n]{n!}} \geq e \sqrt[\leftroot{-2}\uproot{2}n]{\frac{n}{e(n+1)}}$

Taking limit

$\lim_{n \to \infty} e \geq \lim_{n \to \infty} \frac{n}{\sqrt[\leftroot{-2}\uproot{2}n]{n!}} \geq lim_{n \to \infty} \; e \sqrt[\leftroot{-2}\uproot{2}n]{\frac{n}{e(n+1)}}$

Then

$e \geq \lim_{n \to \infty} \frac{n}{\sqrt[\leftroot{-2}\uproot{2}n]{n!}} \geq e$

$\lim_{n \to \infty} \frac{n}{\sqrt[\leftroot{-2}\uproot{2}n]{n!}}=e$