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?
 A: Hint:
$$0<\frac 1{(n!)^\frac 1{n}}=\sqrt[n]{\frac{1}{n!}}<\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}\to 0.$$
In general, we have: if $a_n\to a$, then
$$\lim_{n\to\infty}\frac{a_1+a_2+\cdots+a_n}{n}=a.$$
A: The limit is $0$, and the following very hands-on argument suffices: for any positive integer $k \le n$ we have
$$n! \ge k^{n-k}$$
(we look at $1 \cdot 2 \cdot \dots \cdot n$ and just ignore the terms less than or equal to $k$ and replace all the terms bigger than $k$ with $k$), which gives
$$\sqrt[n]{n!} \ge k^{1 - \frac{k}{n}}.$$
Taking, for example, $k = \lfloor \frac{n}{2} \rfloor$ gives
$$\sqrt[n]{n!} \ge \left\lfloor \frac{n}{2} \right\rfloor^{\frac{1}{2}}$$
so $\lim_{n \to \infty} \sqrt[n]{n!} = \infty$ and $\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0$. Of course much tighter bounds are available, and even quite exact asymptotics via Stirling's formula, but it's worth knowing what can be proven in a very elementary way.
A: $$y=\lim_{n\to\infty}\frac 1{(n!)^\frac 1{n}}$$
$$\ln y=\lim_{n\to\infty}-\frac{1}{n}\ln(n!)$$
$$\ln y=\lim_{n\to\infty}-\frac{1}{n}\left({\ln1+\ln2+\cdots\ln n}\right)$$
$$\ln y=\lim_{n\to\infty}-\frac{1}{n}\left({[\ln\frac{1}{n}+\ln\frac{2}{n}+\cdots\ln\frac{n}{n}]+n\ln n}\right)$$
$$\ln y=\lim_{n\to\infty}\left({\int_0^1\ln x dx-\ln n}\right)$$
$$\ln y=\lim_{n\to\infty}\left({-1-\ln n}\right)\to-\infty$$
$$y=e^{-\infty}\to0$$
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.
