Prove $\lim_{n\to\infty}\frac{1}{n!}^{\frac{1}{n}}=0$ I'm trying to prove that $$\lim_{n\to\infty}\left(\frac{1}{n!}\right)^{\frac{1}{n}}=0$$
Can anyone help me with this?
 A: Think about AM-GM, you get
$(\frac{1}{n!})^{\frac{1}n}\le\frac{2}{n+1}$  
A: An easy way : 
Take , $\displaystyle a_n=\frac{1}{n!}$. Then , $\displaystyle \frac{a_{n+1}}{a_n}=\frac{1}{n+1}\to 0 \text{ , as } n\to \infty $.
So by Cauchy's second limit theorem , $\displaystyle a_n^{1/n}\to 0 \text{ , as } n\to \infty$.
A: We start by writing 
$$\left(\frac1{n!}\right)^{1/n}=e^{-\frac1n \log(n!)}$$
Note that we can easily bound $\log(n!)=\sum_{k=1}^n\log(k)$ using 
$$n\log(n)-n+1=\int_1^n \log(x)\,dx\le \log(n!)\le \int_1^{n+1} \log(x)\,dx=(n+1)\log(n+1)-n$$
Therefore, we have the inequalities
$$-\left(1+\frac1n\right)\log(n+1)+1\le -\frac{\log(n!)}{n}\le -\log(n)+1-\frac1n$$
which obviously show that $\lim_{n\to \infty}-\frac1n\log(n!)= -\infty$.  Finally, we can conclude that 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(\frac1{n!}\right)^{1/n}=0}$$
And we are done!
A: Another way considering $$A_n=\left(\frac{1}{n!}\right)^{\frac{1}{n}}$$ Taking logarithms $$\log(A_n)=-\frac 1n \log(n!)$$ Now, as Shailesh suggested, use Stirling approximation $$\log(n!)\approx n \log(n)-n+\frac 12 \log(2\pi n)$$ which makes $$\log(A_n)\approx -\log(n)+1-\frac 1{2n} \log(2\pi n)$$  So, when $n\to \infty$, $\log(A_n)\to  -\infty$ and $A_n \to 0$.
A: Let $[x]$ denote the largest integer not exceeding $x.\;$ For $n\geq 4$ we have $$[n/2]\geq (n-1)/2 \;\text { and } \quad  n-[n/2]\geq n/2.$$ Therefore, for $n\geq 4$ we have $$n!\geq \prod_{j=1+[n/2]}^n j\geq [n/2]^{(n-[n/2])}\geq ((n-1)/2)^{n/2}>1.$$ Hence $n\geq 4\implies n!^{1/n}\geq ((n-1)/2)^{n/2})^{1/n}=\sqrt {(n-1)/2}.$ 
A: Using Stirling's Approximation:
$$n!\sim\sqrt{2\pi n}(\frac n e)^n \text{ as } n\to\infty$$
We have:
\begin{align}
\lim_{n\to\infty}(\frac 1 {n!})^{\frac 1n}&=\lim_{n\to\infty}[\sqrt{2\pi n}(\frac n e)^n]^{-\frac 1n}\\&=\lim_{n\to\infty}[(2\pi)^{-\frac 1 {2n}}\cdot n^{-\frac 1{2n}}\cdot\frac en]\\&=\lim_{n\to\infty} (1\cdot1\cdot0)\\&=0
\end{align}
