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

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?

## marked as duplicate by Hans Lundmark, Henrik, aes, kingW3, Sahiba AroraAug 4 '17 at 17:33

• Have you heard of the squeeze theorem? – James Oct 6 '14 at 21:40
• yes but how do I get to 2(2/3)^(n-2)>=2^n/n!? – Bono Oct 6 '14 at 21:41
• Like how would I get to the step of 2(2/3)^(n-2)? – Bono Oct 6 '14 at 21:41
• Induction would work. – James Oct 6 '14 at 21:42
• This answer uses exactly the approach from your hint. – Martin Sleziak Apr 17 '17 at 13:43

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.

• how do you get the inequality 2^n/n!<=2(2/3)^(n-2)? – Bono Oct 6 '14 at 21:50
• This inequality is equivalent to $$\frac1{3\times \cdots\times n}\le\frac1{3^{n-2}}$$ which is true since $$3\le k,\quad \forall 3\le k\le n$$ – user63181 Oct 6 '14 at 21:55
• how can you prove that lim(2/3)^(n-2)=0? Using epsilon – Bono Oct 6 '14 at 23:43

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 $a_n=\frac{2^n}{n!}$ ;-)