How do we tackle this limit $\lim_{n\to \infty}2\cdot{\sqrt[n-1]{(n-1)!}\over n}-{\sqrt[n]{n!}\over n-1}={1\over e}?$

Question

Consider the limit $(1)$

$$\lim_{n\to \infty}2\cdot{\sqrt[n-1]{(n-1)!}\over n}-{\sqrt[n]{n!}\over n-1}=L\tag1$$

How does one show that $L={1\over e}$?

I haven't see how this kind of limit is tackle before.

Answer 1

(1) $$\ell_1=\lim_{n\to \infty}\sqrt[n-1]{a_{n}}=\lim_{n\to \infty}\sqrt[n-1]{\frac{(n-1)!}{n^{n-1}}}$$ is the term $n$-th of a series which calculated with Cauchy criterion, that is equivalent with ratio criterion, then

$$\ell_1=\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{\frac{n!}{(n+1)^n}}{\frac{(n-1)!}{n^{n-1}}}=\lim_{n\to \infty}\frac{1}{(1+\frac1n)^n}=\frac1e$$

(2) $$\ell_2=\lim_{n\to \infty}\sqrt[n]{b_{n}}=\lim_{n\to \infty}\sqrt[n]{n!\over(n-1)^n}$$ as (1) is the term $n$-th of a series, with ratio criterion

$$\ell_2=\lim_{n\to \infty}\frac{b_{n+1}}{b_n}=\lim_{n\to \infty}\frac{\frac{(n+1)!}{n^{n+1}}}{\frac{n!}{(n-1)^n}}=\lim_{n\to \infty}\frac{n+1}{n}(1-\frac1n)^n=\frac1e$$

From (1) and (2) the conclusion is $\color{blue}{\dfrac1e}$.

Answer 2

Consider $$A=\sqrt[p]{p!}\implies \log(A)=\frac 1 p \log(p!)$$ and, just as M. Winter commented, use Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+O\left(\frac{1}{p}\right)$$ This makes $$\log(A)=(\log (p)-1)+\frac{\log (2 \pi )+\log \left({p}\right)}{2 p}+O\left(\frac{1}{p^2}\right)$$ Using Taylor again $$A=e^{\log(A)}=\frac{p}{e}+\frac{\log (2 \pi )+\log \left({p}\right)}{2 e}+O\left(\frac{1}{p^2}\right)$$ I am sure that you can take it from here.