Let $\zeta(n)$ denote the Riemann Zeta function defined for positive integers $n$ as usual by:

$$ \zeta(n)=\sum_{m=1}^{\infty} \frac{1}{m^n}. $$

It is currently unknown whether there exists a series representation of $\zeta$ for odd integers $n>3$ of the form:

$$ \zeta (n)=\xi _{{n}}\sum _{{m=1}}^{{\infty }}{\frac {(-1)^{{m-1}}}{m^{{n}}{\binom {2m}{m}}}}, $$

where $\xi_n$ is an algebraic number. For instance, for the case $n=3$, we have $\xi_3=\frac{5}{2}$. If such a $\xi_n$ is found, for any odd $n>3 $, one could use Apéry's proof of the irrationality of $\zeta(3)$ to prove the irrationality of $\zeta(n)$. Despite extensive computer searching, no such $\xi_n$ was found for $n>3$ and it can be shown (1) that if one exists, then the coefficients in its minimal polynomial must be enormous.

You can imagine my surprise when I found that

$$ \xi_7 \approx \frac{\pi^7}{500^2 \sinh^{-1}(\frac{1}{2})^7}, $$

which, despite the fact that it contains a power of $\pi$, is even more surprising given its striking similarity to the well-known value for $\xi_2$:

$$ \xi_2 = \frac{\pi^2}{12 \sinh^{-1}(\frac{1}{2})^2}. $$

I know that seemingly remarkable mathematical coincidences are easy to generate, however this one is far too remarkable to immediately dismiss as pure coincidence. Don't get me wrong, it is most probably a coincidence, but a beatiful one indeed.

How may one attempt to explain this coincidence? Is there a "good" reason for this, other than chance?


(1) D. H. Bailey, J. Borwein, N. Calkin, R. Girgensohn, R. Luke, and V. Moll, Experimental Mathematics in Action, 2007.

  • $\begingroup$ Did you find this result by just guessing values and comparing numerically for a long time or this there a reason you tried this particular value? $\endgroup$ – Václav Mordvinov May 30 '18 at 9:17
  • $\begingroup$ @VáclavMordvinov Intelligent guesswork :) I was trying to find $x$ such that $\xi_n \approx \frac{\pi^n}{x \sinh^{-1}(\frac{1}{2})^n}$. No such "interesting" $x$ exists for odd $n$ except at $n=7$. $\endgroup$ – Klangen May 30 '18 at 9:21
  • $\begingroup$ And how accurate is the approximation of $\xi_2$, and how accurate is the one of $\xi_7$? $\endgroup$ – Václav Mordvinov May 30 '18 at 9:27
  • 2
    $\begingroup$ How can you be surprised when you were specifically looking for this form ? $\endgroup$ – Yves Daoust May 30 '18 at 9:27
  • $\begingroup$ @VáclavMordvinov The value for $\xi_2$ is not an approximation. $\endgroup$ – Klangen May 30 '18 at 9:28

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