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Exercise. Show, without using derivatives, that $\min\{ \frac {\left( 1+\frac 1 n\right)^{n^2}}{2^n} |n\in \Bbb N\} = 1$.

Attempt. I found that when $n=1$, we have the expression equal to $1$. So I decided to show that the expression is an increasing sequence so that it cannot be less than $1$. I tried to use induction, but it has always ended without success. Then, I asked myself when is the expression greater than or equal to $1$, besides $n=1$:

$$\frac {\left( 1+\frac 1 n\right)^{n^2}}{2^n}\geq 1; \iff \left( 1+\frac 1 n\right)^{n^2}\geq2^n; \iff n\ln{\left(1+\frac1n\right)}\geq\ln2; \iff n\ln(n+1)-n\ln n\geq \ln2.$$

Unfortunately, it did not help.

Any opinions shall be appreciated.

Lord Mansfield
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  • Do you know that the sequence $(1+1/n)^n$ monotonically increases? (to $e$). That can be shown using the Bernoulli inequality and some algebraic tricks, and has been done on MSE before. I will try to drag up the relevant post if you want – FShrike Aug 24 '22 at 13:10
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    But, with that, the result is immediate as then you know $(1+1/n)^n\ge2$, so the expression is $\ge2^n/2^n=1$. – FShrike Aug 24 '22 at 13:11
  • Thank you! The problem I have noticed is that $2^n$ also increases, so I would be glad if you could find the post. – Lord Mansfield Aug 24 '22 at 13:13
  • @FShrike But I have $n$ multiplying the logarithm on one side only. – Lord Mansfield Aug 24 '22 at 13:15
  • One such post is this one. I was thinking of a much older post by Robjohn but I couldn’t find it, this newer post mimics the exact same proof anyway – FShrike Aug 24 '22 at 13:15
  • @FShrike But the question also involves that $2^n$ increases too; how do I know that $(1+1/n)^{n^2}$ will not grow slower, at least for some values of $n$? – Lord Mansfield Aug 24 '22 at 13:18
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    @FShrike Ahhhh! I understood what you said. Thank you very, very much!!! – Lord Mansfield Aug 24 '22 at 13:24
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    :) you are welcome – FShrike Aug 24 '22 at 13:35

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