Proof $\lim \limits_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0$ using the exponential series.

Question

I am trying to proof that $$\lim \limits_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0$$

using the exponential series $$E(x)=\sum_{n=0}^\infty \frac{x^{n}}{n!}$$

I am aware that I can accomplish the proof using the Taylor series, AM/GM and other methods but I am looking for an approach that focuses on the exponential series.

However, I am struggling to do so, as I not sure what the connection between the series and sequence would be. The only thing I can think of is that, as a consequence of the series converging, I know that $$\lim \limits_{n \to \infty} \frac{x^{n}}{n!} = 0$$

but I don't know if that is the right track and how I would precede from there...

Any help would be greatly appreciated!

Answer 1

If $x\gt 0$ the series converges, so by the root test we have $$\limsup_{n\to\infty}\sqrt[n]{\frac{x^n}{n!}}= x\limsup_{n\to\infty}\frac{1}{\sqrt[n]{n!}} \le 1 $$

Since $x$ can be arbitrarily large, the $\limsup$ must be zero, which implies the limit is also zero.