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Consider the following, where $n\geq 7$ is a natural number:

$$\frac{1}{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}.$$

I am convinced that there should be an upper bound of the form

$$\frac{1}{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}\leq e^{-c(n)},$$

for $c(n)$ some positive function of $n$ but I am failing dismally to derive one.

Any help would be gratefully appreciated.

Edit: I think $c(n)\approx 4.689$ works and suffices for my needs. This is independent of $n$.

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Note that we can write

$$\begin{align} \frac1{2^{n^2}}\left(1+\frac{1}{e^{\pi^2/n^2}}\right)^{n^2}&=e^{-\pi^2}\left(1+\frac{e^{\pi^2/n^2}-1}{2}\right)^{n^2}\\\\ &\le e^{-c(n)} \end{align}$$

where $c(n)$ is given by

$$c(n)\le\pi^2-n^2\log\left(1+\frac{e^{\pi^2/n^2}-1}{2}\right)$$


The asymptotic expansion of $c(n)$ is given by

$$\begin{align} c(n)&=\pi^2-n^2 \log\left(1+\frac{(\pi^2/n^2+\frac12(\pi^2/n^2)^2+\frac16(\pi^2/n^2)^3+O\left(1/n^8\right)}{2}\right)\\\\ &=\pi^2-n^2\left(\frac{\pi^2}{n^2}+\frac{\pi^4}{8n^4}+O(1/n^8)\right)\\\\ &=\frac{\pi^2}{2}-\frac{\pi^4}{8n^2}+O\left(\frac1{n^6}\right)\\\\ &=c_{\text{approx}}+O\left(\frac1{n^6}\right) \end{align}$$

where $c_{\text{approx}}=\frac{\pi^2}{2}-\frac{\pi^4}{8n^2}$.


For $n=7$, we have

$$\begin{align}c(7) &\approx 4.68672854698357\\ c_{\text{approx}}(7) &=4.68630962137631 \end{align}$$


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  • $\begingroup$ Mark: glad to see 4.68 reappear! Thank you so much for this answer. $\endgroup$ – JP McCarthy May 31 '18 at 15:50
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    $\begingroup$ @JpMcCarthy You're welcome. My pleasure. The zero order approximation is $\pi^2/2\approx 4.93480220054468$. But this overestimates $c(n)$ for all $n$. $\endgroup$ – Mark Viola May 31 '18 at 16:03

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