Prove that $\lim_{n \rightarrow \infty} \frac{2^{n}}{n!} = 0$

Question

Prove that $\lim_{n \rightarrow \infty} \frac{2^{n}}{n!} = 0$ using the hint that $0 < \frac{2^{n}}{n!} \leq 2 \, \left(\frac{2}{3}\right)^{n-2}$ for all $n \geq 3$? I know there is a thread already about this question but there is none that use this hint to solve this question. How can you use this hint to prove it?

Answer 1

The inequality is trivial for $n=3$. For the inductive step:

$$\frac{2^{n+1}}{(n+1)!}=\frac2{n+1}\frac{2^n}{n!}\le \frac2{n+1}\times 2\left(\frac23\right)^{n-2}\le2 \left(\frac23\right)^{n-1}$$ since $n+1\ge3$.

Now to use the hint notice that the geometric sequence $\left(\frac23\right)^{n-2}$ is convergent to $0$ so we conclude the desired result using the squeeze theorem.

Answer 2

Hint: If $$\lim_{n\to\infty }\left|\frac{a_{n+1}}{a_n}\right|=\rho<1,$$ by d'Alembert rule, $$\lim_{n\to\infty }a_n=0.$$

I let you try for $a_n=\frac{2^n}{n!}$ ;-)