This question already has an answer here:

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 Arora Aug 4 '17 at 17:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Have you heard of the squeeze theorem? $\endgroup$ – James Oct 6 '14 at 21:40
  • $\begingroup$ yes but how do I get to 2(2/3)^(n-2)>=2^n/n!? $\endgroup$ – Bono Oct 6 '14 at 21:41
  • $\begingroup$ Like how would I get to the step of 2(2/3)^(n-2)? $\endgroup$ – Bono Oct 6 '14 at 21:41
  • $\begingroup$ Induction would work. $\endgroup$ – James Oct 6 '14 at 21:42
  • 1
    $\begingroup$ This answer uses exactly the approach from your hint. $\endgroup$ – 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.

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


Not the answer you're looking for? Browse other questions tagged or ask your own question.