 # Proving that \frac{\pi^3}{32}=1-\sum_{k=1}^\infty\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} spiderifilms6e 2022-01-22 Answered
Proving that $\frac{{\pi }^{3}}{32}=1-\sum _{k=1}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)\zeta \left(2k+2\right)}{{4}^{2k+2}}$
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Yes, we can prove it. We can change the order of summation in
$\sum _{k=1}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)\zeta \left(2k+2\right)}{{4}^{2k+2}}=\sum _{k=1}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)}{{4}^{2k+2}}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2k+2}}$
$\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)}{{\left(4n\right)}^{2k+2}}$
$\sum _{n=1}^{\mathrm{\infty }}r\left(4n\right)$
where, for $|z|>1$, we define
$r\left(z\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{z}^{2k}}=\frac{1}{{z}^{2}-1}=\frac{1}{2}\left(\frac{1}{z-1}-\frac{1}{z+1}\right)$
Differentiating yields $r\left(z\right)=\frac{1}{{\left(z-1\right)}^{3}}-\frac{1}{{\left(z+1\right)}^{3}}$, so
$1-\sum _{k=1}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)\zeta \left(2k+2\right)}{{4}^{k+2}}=\sum _{v=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{v}}{{\left(2v+1\right)}^{3}}$
Using the partial fraction decomposition of $\frac{1}{\mathrm{cos}z}$:
$\sum _{v=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{v}}{{\left(2v+1\right)}^{3}}=-\frac{{\pi }^{3}}{32}{E}_{2}=\frac{{\pi }^{3}}{32}$
###### Not exactly what you’re looking for? nebajcioz
Starting from the Laurent series of the cotangent function:
$\pi z\mathrm{cot}\left(\pi z\right)=1-2\sum _{k=0}^{\mathrm{\infty }}\zeta \left(2k+2\right){z}^{2k+2}$
apply the differential operator:
$\stackrel{^}{D}={z}^{2}\frac{{d}^{2}}{{dz}^{2}}-2z\frac{d}{dz}+2$
to get:
${z}^{3}{\pi }^{3}\mathrm{cot}\left(\pi z\right)\left(1+{\mathrm{cot}\left(\pi z\right)}^{2}\right)=1-\sum _{k=0}^{\mathrm{\infty }}2k\left(2k+1\right)\zeta \left(2k+2\right){z}^{2k+2}$
which, by the ratio test, has a radius of convergence of $|z|<1$. Then from:

we have:
$\frac{{\pi }^{3}}{32}=1-\sum _{k=0}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)\zeta \left(2k+2\right)}{{4}^{2k+2}}$
###### Not exactly what you’re looking for? RizerMix
$S=\sum _{k=1}^{\mathrm{\infty }}\frac{2k\left(2k+1\right)\zeta \left(2k+2\right)}{{4}^{2k+2}}=\sum _{k=1}^{\mathrm{\infty }}\frac{\left(2k+1\right)!\zeta \left(2k+2\right)}{\left(2k-1\right)!{4}^{2k+2}}$ $=\sum _{k=1}^{\mathrm{\infty }}\frac{\left(1/4{\right)}^{2k+2}}{\left(2k-1\right)!}{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2k+1}}{{e}^{x}-1}dx={4}^{-3}{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}}{{e}^{x}-1}\left(\sum _{k=1}^{\mathrm{\infty }}\frac{\left(x/4{\right)}^{2k-1}}{\left(2k-1\right)!}\right)dx$ $={4}^{-3}{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}}{{e}^{x}-1}\mathrm{sinh}\left(\frac{x}{4}\right)dx=\frac{{4}^{-3}}{2}{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}}{1-{e}^{-x}}\left({e}^{-\frac{3}{4}x}-{e}^{-\frac{5}{4}x}\right)dx$ $={4}^{-3}\left[\zeta \left(3,\frac{3}{4}\right)-\zeta \left(3,\frac{5}{4}\right)\right]=1-{4}^{-3}\left[\zeta \left(3,\frac{1}{4}-\zeta \left(3,\frac{3}{4}\right)\right]$ $=1-\beta \left(3\right)=1-\frac{{\pi }^{3}}{32}$