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

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 , $lim_{n\to\infty} \frac{1}{(a_n)^{1/n}}=0$
