COMPUTING SOME INFINITE SUMS WITH THE AID OF POLYGAMMA FUNCTION PART 2

        In previous post  we computed the third and fourth infinite sum from the list below. Today we will proof the first and last two.


\begin{aligned} &\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=\frac{117 \zeta(3)+2 \sqrt{3} \pi^{3}}{243} \\ &\sum_{n=0}^{\infty} \frac{1}{(3 n+2)^{3}}=\frac{117 \zeta(3)-2 \sqrt{3} \pi^{3}}{243} \\ &\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}=\frac{28 \zeta(3)+\pi^{3}}{64} \\ &\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}}=\frac{28 \zeta(3)-\pi^{3}}{64} \\ &\sum_{n=0}^{\infty} \frac{1}{(6 n+1)^{3}}=\frac{91 \zeta(3)+2 \sqrt{3} \pi^{3}}{216} \\ &\sum_{n=0}^{\infty} \frac{1}{(6 n+5)^{3}}=\frac{91 \zeta(3)-2 \sqrt{3} \pi^{3}}{216} \end{aligned}

\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=\frac{117 \zeta(3)+2 \sqrt{3} \pi^{3}}{243}

Proof:

\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=\frac{1}{3^3}\sum_{n=0}^{\infty} \frac{1}{\left( n+\frac{1}{3}\right)^{3}}

Recall that (which we proved here)

\psi^{\prime \prime}\left(x\right)=-2\sum_{n=0}^{\infty} \frac{1}{(n +x)^{3}}

Then

\sum_{n=0}^{\infty} \frac{1}{(n +x)^{3}}=-\frac{\psi^{\prime \prime}\left(x\right)}{2}

Letting    x=\frac{1}{3}  we get

\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=-\frac{1}{3^3} \cdot \frac{\psi^{\prime \prime}\left(\frac{1}{3}\right)}{2}

From here we know that

\psi^{\prime \prime}\left(\frac{1}{3}\right)=-\frac{4\pi^3}{3 \sqrt{3}}-26\zeta(3)

\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=\frac{2\pi^3}{81 \sqrt{3}}+\frac{13\zeta(3)}{27}

\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=\frac{2 \sqrt{3} \pi^3}{243 }+\frac{13\zeta(3)}{27}

\boxed{\sum_{n=0}^{\infty} \frac{1}{(3 n+1)^{3}}=\frac{117 \zeta(3)+2 \sqrt{3} \pi^{3}}{243}}

Similarly

\sum_{n=0}^{\infty} \frac{1}{(3 n+2)^{3}}=\frac{117 \zeta(3)-2 \sqrt{3} \pi^{3}}{243}

\sum_{n=0}^{\infty} \frac{1}{(3 n+2)^{3}}=\frac{1}{27}\sum_{n=0}^{\infty} \frac{1}{\left( n+\frac{2}{3}\right)^{3}}

\sum_{n=0}^{\infty} \frac{1}{(3 n+2)^{3}}=-\frac{1}{27}\frac{\psi^{\prime \prime}\left(\frac{2}{3}\right)}{2}

From here we know that

\psi^{\prime \prime}\left(\frac{2}{3}\right)=\frac{4\pi^3}{3 \sqrt{3}}-26\zeta(3)

And

\boxed{\sum_{n=0}^{\infty} \frac{1}{(3 n+2)^{3}}=\frac{117 \zeta(3)-2 \sqrt{3} \pi^{3}}{243}}

\sum_{n=0}^{\infty} \frac{1}{(6 n+1)^{3}}=\frac{91 \zeta(3)+2 \sqrt{3} \pi^{3}}{216}

\sum_{n=0}^{\infty} \frac{1}{(6 n+1)^{3}}=\frac{1}{6^3}\sum_{n=0}^{\infty} \frac{1}{\left( n+\frac{1}{6}\right)^{3}}

\sum_{n=0}^{\infty} \frac{1}{(6 n+1)^{3}}=-\frac{1}{216}\frac{\psi^{\prime \prime}\left(\frac{1}{6}\right)}{2}

From here we know that

\psi^{\prime \prime}\left(\frac{1}{6}\right)=-182 \zeta(3)-4 \sqrt{3} \pi^3

Plugging it in the equation above gives us the desired result

\boxed{\sum_{n=0}^{\infty} \frac{1}{(6 n+1)^{3}}=\frac{91 \zeta(3)+2 \sqrt{3} \pi^{3}}{216} }

Finally, following the same steps as before and using the fact that

\psi^{\prime \prime}\left(\frac{5}{6}\right)=-182 \zeta(3)+4 \sqrt{3} \pi^3

we obtain

\boxed{\sum_{n=0}^{\infty} \frac{1}{(6 n+5)^{3}}=\frac{91 \zeta(3)-2 \sqrt{3} \pi^{3}}{216}}

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