Prove $\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0$ I used $$(n!)^{\frac{1}{n}}=e^{\frac{1}{n}\ln(n!)}=e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}$$
Then using Stirling's approximation and L'Hospital's rule on $$\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}$$ I get $$\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}=\lim\limits_{n\to\infty}(\ln(n)+\frac{n+\frac{1}{2}}{n}-1)=\infty$$
Now, $$e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}=e^{\infty}=\infty$$
Thus
$$\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]n}=\frac{1}{\infty}=0$$
Is this correct approach and what other methods could be used?
 A: As long as you haven't made an algebra mistake, stirlings approximation should work.
Separate $n!$ into two parts. Assign $1$ to everything below $n/2$. Assign $n/2$ to everything above $n/2$
So you get $(n/2)^{n/2}<n!$
Apply the root and you get:
$(n/2)^{1/2}<n!^{1/n}$
$(n/2)^{1/2}$ clearly aproaches infinity so it will make the limit zero.
A: Since $\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=m$, we have:
$$ n!=\prod_{m=2}^{n}m = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}$$
so:
$$ n!\geq \frac{n^n}{e^{n-1}} $$
and the claim easily follows.
A: You don't even have to use l'Hopital's rule; you can just plug in Sterling's formula and divide by $n$, then take limits.
Another way would be to use arithmetic-geometric means:
$${1 \over (n!)^{1 \over n}} = (\prod_{k=1}^n {1 \over k})^{1 \over n}
\leq {1 \over n}\sum_{k = 1}^n {1 \over k}$$
Since $\sum_{k = 1}^n {1 \over k}$ grows as $\ln n$ the limit is zero.
A: If we can prove that if $\left | \frac{a_{n+1}}{a_{n}} \right |\rightarrow L$, then $a_{n}^{1/n}\rightarrow L$, we may simply take $a_{n}=n!$ to get the result. 
So suppose that $\left | \frac{a_{n+1}}{a_{n}} \right |\rightarrow L$. Then there is an integer $N$ such that if $n>N$, then 
($^{*}$) $L-\epsilon <\left | \frac{a_{n+1}}{a_{n}} \right |<L+\epsilon $.
Now, for $n>N$, 
$\left | a_{n} \right |=\left | \frac{a_{n}}{a_{n-1}} \right |\left | \frac{a_{n-1}}{a_{n-2}} \right |\left | \frac{a_{n-2}}{a_{n-3}} \right |\cdots \left | \frac{a_{N+1}}{a_{N}} \right |\left | a_{N} \right |$ 
and so if we appeal to ($^{*}$), we see that this is less than 
$(L+\epsilon )^{n-N}\left | a_{N} \right |$ 
and greater than 
$(L-\epsilon )^{n-N}\left | a_{N} \right |$
Now take the $n$th root to obtain
$\frac{L-\epsilon}{(L-\epsilon)^{N/n}}a_{N}^{1/n}<a_{n}^{1/n}<\frac{L+\epsilon}{(L+\epsilon)^{N/n}}a_{N}^{1/n}$. Letting $n\rightarrow \infty $, we get the result. 
A: Using Stirling's Approximation: 
$$n!\sim\sqrt{2\pi n }(\frac{n}{e})^n$$
Also 
$$\lim\limits_{k\to\infty}k^{\frac1k}=1$$
So we have
\begin{align}
\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]n}&=\lim\limits_{n\to\infty}\frac1{[\sqrt{2\pi n }(\frac{n}{e})^n]^{\frac{1}n}}=\lim\limits_{n\to\infty}\frac1{\frac{n}e\cdot(2\pi n)^{\frac1{2n}}}\\&=\lim\limits_{n\to\infty}\frac e{n\cdot\pi^{\frac1{2n}}\cdot(2n)^{\frac1{2n}}}\\&=\lim\limits_{n\to\infty}\frac{e}{n}\\&=0 
\end{align}
A: $$\ln [(n!)^{1/n}]= (1/n)\sum_{k=1}^{n}\ln k \ge (1/n)\sum_{k>n/2}\ln k \ge (1/n)[(n/2)\ln (n/2)] = (1/2)\ln (n/2) \to \infty.$$ Thus $(n!)^{1/n} \to \infty$ and the desired limit is $0.$
A: We have
$$\newcommand{\limti}[1]{\lim\limits_{#1\to\infty}}0\le \limti n \frac1{\sqrt[n]{n!}} = \limti n \sqrt[n]{1\cdot\frac12\cdots\frac1n} \overset{(1)}\le \limti n \frac{1+\frac12+\dots+\frac1n}n \overset{(2)}=0.$$
(1) is AM-GM inequality
(2) is special case of the result that
$$\limti n a_n = L \qquad\Rightarrow\qquad \limti n \frac{a_1+\dots+a_n}n=L.$$
See Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means, If $a_n\to \ell $ then $\hat a_n\to \ell$ and A result on sequences: $x_n\to x$ implies $\frac{x_1+\dots+x_n}n\to x$ without using Stolz-Cesaro.
