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

Question

show that $$\lim_{n\to\infty}\dfrac{1}{\sqrt[n]{n!}}=0$$

I want see more methods and if someone have nice methods,

Thank you

Answer 1

We have $$e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!} + \cdots > \dfrac{x^n}{n!}$$ Plugging in $x=n$, we get that $$e^n > \dfrac{n^n}{n!} \implies e > \dfrac{n}{\sqrt[n]{n!}} \implies 0 < \dfrac1{\sqrt[n]{n!}} < \dfrac{e}n$$ Now take the limit as $n \to \infty$ and use Sandwich/Squeeze lemma.

Answer 2

Prove that $(n!)^{1/n}\to\infty$ as follows.

Take $A>0$. Choose $k>A$. Then for $n$ large enough $$n\cdot n-1\cdot\;\cdots\;\cdot k+1\cdot k\cdot\;\cdots\;\cdot 2 \cdot 1>\underbrace{n\cdot n-1\cdot\;\dots\;\cdot k+1}_{n-k\text{ terms }}>(k+1)^{n-k}$$

So that $$n!^{1/n}>(k+1)^{1-k/n}$$

Thus $$\limsup_{n\to\infty}\left(n!^{1/n}\right)>k+1>A$$

Answer 3

This would be a comment, not an answer, but I don't have the rep. You might be interested in Stirling's approximation, which says that $n!$ is approximately $(\frac{n}{e})^n$. Then your limit term is (approximately) equal to $\frac{e}{n}$, which clearly goes to 0.

Depending on your background and goals, you may want to spend some time with the precise statement of the approximation (check Wikipedia for more information than you could possibly want) and prove this more rigorously, with a "squeeze theorem" type argument.

Answer 4

The easiest method I can think of would be the following:

Take the factorial, and multiply in this pattern:

$$ n!=(1\cdot n)\cdot(2\cdot(n-1))\cdot(3\cdot(n-2))\cdots $$

Now, for even $n>2$, we can see that all of the terms except the first are greater than $n$. For odd $n>2$, the final term is $n/2>1$. Therefore, we can say with confidence that $n! > n^{\lfloor n/2\rfloor}$. Thus,

$$ (n!)^{1/n} > n^{(n/2-1)/n} = n^{1/2-1/n} $$ and so $$ \lim_{n\to\infty} \sqrt[n]{n!} \geq \lim_{n\to\infty} n^{1/2-1/n} \to \infty $$ The final step is then obvious.