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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?

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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
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    $\begingroup$ This answer uses exactly the approach from your hint. $\endgroup$ – Martin Sleziak Apr 17 '17 at 13:43
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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.

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

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