find the limit of this function? Evaluate $$\lim_{n\to \infty}\frac{1}{(n!)^{1/n}}={?}$$
My try: 
$$\lim_{n\to \infty}\frac{1}{(n!)^{1/n}}=\exp{\lim_{n\to \infty}\frac{-\ln(n!)}{n}}$$
$$=\exp{\lim_{n\to \infty}\frac{-\ln(1\times2\times3\ldots(n-1)\times n)}{n}}$$
$$=\exp{\lim_{n\to \infty}\frac{-(1+2+3+\ldots+(n-1)+n)}{n}}$$
$$=\exp{(-\lim_{n\to \infty}\left(\frac1n+\frac2n+\ldots+\frac{n-1}{n}+\frac nn\right)})$$
$$=\exp(-1)=\frac1e$$
I don't know if my answer is correct please correct me if i am wrong. thanks
 A: Note that $e^x \geq \frac{x^n}{n!}$ for all $x\geq 0$. Setting $x=n$, we have $$ n! \geq \left( \frac{n}{e} \right)^n .$$
Thus $$(n!)^{1/n} \geq \frac{n}{e} \to \infty.$$
A: You were almost there! Remember that: $$1 + 2 + \cdots + n = \frac{n(n+1)}{2}.$$ If you use this, then you will see that the expression inside $\operatorname{exp}$ actually goes to $- \infty$.
A: Use Cauchy's formula on limits:
let $a_n=n!$
$lim_{n\to\infty} \frac{a_{n+1}}{a_n}=lim_{n\to\infty} (n+1)=\infty>1$
$\implies lim_{n\to\infty} (a_n)^{1/n}=\infty$
so , $lim_{n\to\infty} \frac{1}{(a_n)^{1/n}}=0$
A: Since
$$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\ell\implies \lim_{n\to \infty} \sqrt[n] a_n=\ell$$
Let
$$a_n=\frac{1}{n!}$$
then
$$\frac{a_{n+1}}{a_n}=\frac{n!}{(n+1)!}=\frac{1}{n+1}\to0 \implies \lim_{n\to \infty}\frac{1}{(n!)^{1/n}}=0$$
A: $$\frac{1}{(n!)^{1/n}}=e^{-\frac1n \log n!}\to e^{-\infty}=0$$
indeed by Stolz-Cesaro:
$$\lim_{n\to \infty} \frac1n \log n!=\lim_{n\to \infty}\frac{\log (n+1)!-\log n!}{n+1-n}=\lim_{n\to \infty}\log (n+1)=+\infty$$
A: 
Stirling's approximation $$n!\approx\sqrt{2\pi n}\left(\dfrac ne\right)^n$$

$$\begin{align}L&=\lim_\limits{n\to\infty}\dfrac1{(n!)^{\frac1n}}\\&=\lim_\limits{n\to\infty}\dfrac1{(2\pi n)^{\frac 1{2n}}\cdot\dfrac ne}\\&=\lim_\limits{n\to\infty}\dfrac{e}{ne^{\frac{\ln(2\pi n)}{2n}}}\\&=\lim_\limits{n\to\infty}\dfrac e{ne^{\frac1{2n}}}\qquad\left[\text{Applying L' H}\hat{o}\text{pital's Rule on }\lim_\limits{n\to\infty}\dfrac{\ln(2\pi n)}{2n}\right]\\&=\lim_\limits{n\to\infty}\dfrac en\\&=0\end{align}$$
