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?

Solution 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.

Solution 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.

Solution 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.