Need to find $\lim_{n\to\infty}$ $\frac 1{(n!)^\frac 1{n}}$ I've attempted to solve this  $\lim_{n\to\infty}$ $\frac 1{(n!)^\frac 1{n}}$ via converging 1/n to zero, but somehow it's incorrect. Can someone explain what i'm doing wrong?
Hint:
$$0&lt;\frac 1{(n!)^\frac 1{n}}=\sqrt[n]{\frac{1}{n!}}&lt;\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}\to 0.$$
In general, we have: if $a_n\to a$, then
$$\lim_{n\to\infty}\frac{a_1+a_2+\cdots+a_n}{n}=a.$$
Use the common limit
$$\lim_{n \to \infty} \left( \frac{n^n}{n!}\right)^\frac 1n = \lim_{n \to \infty} n \times \frac{1}{(n!)^\frac1n} = e$$
It's easy to see that $$\lim_{n \to \infty} \frac{1}{(n!)^\frac1n} = \lim_{n \to \infty} \frac{e}{n} = 0$$
And that's the final answer.

It's easier to do this via Stirling's approximation, as $n! = \sqrt{2\pi n}\left( \frac ne\right)^n$ and the nth root of $1/n!$ is equal to $e/n$, which tends to 0.