Prove $\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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}

Answer 6

$$\ln [(n!)^{1/n}]= (1/n)\sum_{k=1}^{n}\ln k \ge (1/n)\sum_{k>n/2}\ln k \ge (1/n)[(n/2)\ln (n/2)] = (1/2)\ln (n/2) \to \infty.$$ Thus $(n!)^{1/n} \to \infty$ and the desired limit is $0.$

Answer 7

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.