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Is anything known about the value of $$\zeta(3/2)=\sum_{n\geq 1}\frac{1}{n^{3/2}}?$$ It is a classical result that $\displaystyle \zeta(2)= \frac{\pi^2}{6}$ and $\zeta(3)$ has been shown to be irrational by Roger Apéry in 1979.

Do we even know if $$\zeta(3/2)=\sum_{n\geq 1}\frac{1}{n^{3/2}}$$ is an irrational number or not? Is it true that $$\zeta(3/2)=\sum_{n\geq 1}\frac{1}{n^{3/2}}=\frac{a}{b}\sum_{n\geq 1}\frac{(-1)^{k-1}}{n^{3/2}\binom{2n}{n}}$$ for some $a,b \in \mathbb{Z}$? Not sure if Apéry's proof can be adapted here since I haven't read it.

Links given by J.G in $\zeta(3/2)$ mention that Alcover mentions that $$\zeta(3/2)=\frac{2}{\sqrt{\pi}}\int_0^{\infty}\frac{\sqrt{x}}{e^x-1} dx$$ (using the Mellin transform with the function $\displaystyle f(x)=\frac{1}{e^x-1})$ and Luschny noting that from the functional equation of the Riemann zeta function we have: $$\zeta(3/2)=\Gamma(-1/2)\zeta(-1/2)\tau(-1/2)$$ where $$\tau(s) = (2\pi i)^{-s} + (-2 \pi i)^{-s}$$

Very cool that this is related to the temperature of a Bose-Einstein condensate. Possibly also related to: $\zeta(3)$ in terms of $\zeta'(1/2)$ As a matter of fact, from Juan's answer there, we have that $$2\sum_{n=1}^\infty \frac{1}{(4n+1)^{3/2}}=-2+L(3/2,\chi)+(1-2^{-3/2})\zeta(3/2)$$ hence $$\zeta(3/2)= \frac{1}{\sqrt{2}}\sum_{n=1}^\infty \frac{1}{(4n+1)^{3/2}} +\frac{1}{\sqrt{2}}-\frac{1}{2\sqrt{2}}L(3/2, \chi)$$ where $L(3/2,\chi)$ is the Dirichlet zeta-function at $3/2$.

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    $\begingroup$ Next to nothing is known about values of zeta at odd integers, let alone noninteger elements. I highly doubt anything is really known about this value. $\endgroup$
    – Wojowu
    Commented Nov 26, 2020 at 21:56
  • $\begingroup$ @Wojowu yep my man, nothing at all is known whatsoever about its value! I just recently saw on the wiki it's related to the critical temperature of a Bose-Einstein condensate though, pretty neat. At any rate, even worse, you can't even use any tricks with Parseval-Plancherel and Fourier Series to even take a crack at this problem... $\endgroup$ Commented Nov 26, 2020 at 22:00
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    $\begingroup$ "Not sure if Apéry's proof can be adapted here since I haven't read it." So, what are you waiting for? $\endgroup$ Commented Nov 26, 2020 at 22:18
  • $\begingroup$ @GerryMyerson Thanks for the encouragement, much appreciated! Definitely should read it and see what ingredients go into it! $\endgroup$ Commented Nov 26, 2020 at 22:20

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