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


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 $n$.


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


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

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

  • $\begingroup$ Mark: glad to see 4.68 reappear! Thank you so much for this answer. $\endgroup$ – JP McCarthy May 31 '18 at 15:50
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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