Please prove: $ \lim_{n\to \infty}\sqrt[n]{\frac{1}{n!}} = 0 $ [duplicate]

Question

Possible Duplicate:
$ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

Please prove: $$ \lim_{n\to \infty}\sqrt[n]{\frac{1}{n!}} = 0 $$

Answer 1

$$0<\sqrt[n]{\frac{1}{n!}}=\left(1\cdot\frac{1}{2}\cdots\frac{1}{n}\right)^{\frac{1}{n}}\leq\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}<\frac{1+\ln n}{n}$$

As $\lim_{n\rightarrow\infty}\frac{1+\ln n}{n}=0$, so $\lim_{n\rightarrow\infty}\sqrt[n]{\frac{1}{n!}}=0$

Answer 2

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 $n! \ge \lceil \frac n2\rceil^{\lfloor \frac n2\rfloor}$.

Answer 3

Suppose it is not true. Then we have $\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{1}{n!}}=\frac{1}{x}$ for some $x\geq 1$

By the hypothesis, the limit is not $0$, so we can write it like this: $\displaystyle\lim_{n\to\infty}\sqrt[n]{n!}=x$. Thus, as $n$ goes to infinity, then $n!\to x^n$.

This is clearly a contradiction, because for all $n>x$ the left side ($n!$) increases far more then the right side ($x^n$).

EDIT: i now see that a similar proof has been posted.

EDIT 2: I have'nt shown yet that the limit does exist.

I'll show that the function $f(n)=\sqrt[n]{\frac{1}{n!}}$ is strictly descending:

We have that $n!<(n+1)^n$. (because $n>0$)

Multiplicate bot sides with $(n!)^n$ and we have $(n!)^{n+1}<((n+1)!)^n$. Take the $n*(n+1)$-th root of both sides gives $\sqrt[n]{n!}<\sqrt[n+1]{(n+1)!}$.

Thus, $\sqrt[n]{\frac{1}{n!}}>\sqrt[n+1]{\frac{1}{(n+1)!}}$.

Now, because $f(n)>0$ and $f$ is strictly descending, the function converges and the limit exists.

Answer 4

Just try to put n equal to infinity. since 'n' appears in the denominator it will tend to zero.

Please confirm the answer.