Possible Duplicate:
$ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite
Please prove: $ \lim_{n\to \infty}\sqrt[n]{\frac{1}{n!}} = 0 $
Possible Duplicate:
$ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite
Please prove: $ \lim_{n\to \infty}\sqrt[n]{\frac{1}{n!}} = 0 $
Just try to put n equal to infinity. since 'n' appears in the denominator it will tend to zero.
Please confirm the answer.
Hint: When writing out $n!$, you have \[ n! = n \cdot (n-1) \cdots \left\lceil \frac n2\right\rceil \cdots 1 \] so at least $\lfloor \frac n2\rfloor$ of the factors are larger then $\lceil \frac n2\rceil$. So $n! \ge \lceil \frac n2\rceil^{\lfloor \frac n2\rfloor}$.
$0<\sqrt[n]{\frac{1}{n!}}=\left(1\cdot\frac{1}{2}\cdots\frac{1}{n}\right)^{\frac{1}{n}}\leq\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}<\frac{1+\ln n}{n}$
As $\lim_{n\rightarrow\infty}\frac{1+\ln n}{n}=0$, so $\lim_{n\rightarrow\infty}\sqrt[n]{\frac{1}{n!}}=0$