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I came across a problem with proving an equality:

$n! < e(\frac{n}{2})^n$

I tried to simplify the RHS using the limit of $(1+\frac{1}{n})^n$ as $n$ goes to infinity (to use with $sandwich$), but to no avail. I also thought about using induction, but couldn't prove the inductive step.

ItamarG3
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2 Answers2

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Induction works easily. It is true for $n=1$. The induction step goes as follows:

$(n+1)! = (n+1)n! < (n+1)e(\frac{n}{2})^n$ which is less than $e(\frac{n+1}{2})^{n+1}$ if and only if $2n^n < (n+1)^n \iff 2 < (1+\frac{1}{n})^n$ which is true by Binomial Theorem.

mathworker21
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  • Oh... I didn't think about using the induction assumption right in the beginning... I feel quite foolish now :P

    I'll try it now (Writing it formally, because that seems to be the only way I understand what I'm doing)

    Thanks!!

     – ItamarG3 Dec 18 '16 at 07:15
  • Wait... actually, I'm not sure if I understand... I'm quite new to all these things (Previously I only used function analysis, and not limits and sequences) So could you please explain the steps you made? – ItamarG3 Dec 18 '16 at 07:21
  • which step do you want me to explain – mathworker21 Dec 18 '16 at 08:55
  • no nevermind. I got it – ItamarG3 Dec 18 '16 at 08:58
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Since $\left(1+\frac1n\right)^{n+1}$ is decreasing, we have $$ \begin{align} \frac{2^nn!/n^n}{2^{n+1}(n+1)!/(n+1)^{n+1}} &=\frac12\left(1+\frac1n\right)^n\\ &\ge\frac{en/2}{n+1} \end{align} $$ Therefore, $$ \begin{align} \frac{n!}{n^n} &\le\frac{n}{e^{n-1}}\\ &\le\frac{e}{2^n} \end{align} $$ where the last inequality is valid for $n\ge6$. We can verify $0\le n\lt6$ manually.

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robjohn
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