It is known that $\zeta(2)=\frac{\pi^{2}}{6}$ and that $\zeta(4)=\frac{\pi^{4}}{90}$. Thus, for $\zeta(2n)$, this can be generalized to : $$ \zeta(2n)=\frac{(-1)^{n-1}B_{2n}(2\pi)^{2n}}{2(2n)!} $$ Where $B_{2n}$ are the Bernoulli's numbers. Now why is that such closed form expression does not exist for all $\zeta(2n+1)$?
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3It's not known whether it exists or not. It's not even known the exact value of $\zeta(3)$ for example. – Peanut Sep 28 '20 at 12:11
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https://math.stackexchange.com/questions/12815/riemann-zeta-function-at-odd-positive-integers – Sep 28 '20 at 12:16
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1See also this post. – Dietrich Burde Sep 28 '20 at 12:18
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There was an /r/math post recently about this because someone uploaded a paper to arXiv claiming new results on the sum of inverse cubes (though they were known for at least 20 years IIRC). It created a bit of hubbub from people not realizing it wasn't a real step forward in the solution of this problem. – Cameron Williams Sep 28 '20 at 12:44
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In general, it is an open conjecture whether or not $\zeta(2n+1)$ is a rational multiple of $\pi^{2n+1}$. This not even clear for $n=1$. The following series representation was found by Ramanujan: $$ \zeta (3)=\frac {7}{180}\pi ^{3}-2\sum _{k=1}^{\infty }{\frac {1}{k^{3}(e^{2\pi k}-1)}}. $$ His formula for $\zeta(2n+1)$ can be found here.
Dietrich Burde
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3To add to this, IIRC, if $\zeta(3) = \frac{p}{q} \pi^3$, $p$ and $q$ are extremely large. – Cameron Williams Sep 28 '20 at 12:39
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