How to check if $\sum_{n=1}^\infty \frac 1{n!}$ is converging or diverging by direct comparison test?

Question

$$\sum_{n=1}^\infty \frac 1{n!}$$

I think we may compare it with $\frac{1}{n^n}$, and say that $\frac{1}{n!}>\frac{1}{n^n}$. But I am not sure of the convergence or divergence of $\frac{1}{n^n}$ itself, so this turns out to be useless.

What other way is there?

Answer 1

Compare it to $\dfrac{1}{n^2}$

and for $n\ge4$, we have $$\dfrac{1}{n!}<\dfrac{1}{n^2}$$

Answer 2

hint

By induction, you can prove that for $$n\ge 1, \;\; n!\ge 2^{n-1}$$

thus $$\frac{1}{n!}\le \frac{1}{2^{n-1}}$$