Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$
I don't have an idea about how to start.
 A: What we have is
$$\frac1{1^3} - \frac1{7^3} + \frac1{9^3} - \frac1{15^3} + \cdots $$
Imagine if we used negative summation indices (e.g., $n=0, -1, -2, \cdots$).  Then we would have
$$\frac1{(-7)^3} - \frac1{(-1)^3} + \frac1{(-15)^3} - \frac1{(-9)^3} + \cdots$$
You should see then that the sum is
$$\frac12 \sum_{n=-\infty}^{\infty} \left [\frac1{(8 n-7)^3} - \frac1{(8 n-1)^3} \right ] $$
The significance of this is that we may use a very simple result from residue theory to evaluate the sum:
$$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} [f(z) \cot{\pi z} ]$$
where the $z_k$ are the non-integer poles of $f$.  Here
$$f(z) = \frac12 \left [\frac1{(8 z-7)^3} - \frac1{(8 z-1)^3} \right ] = \frac1{2 \cdot 8^3} \left [\frac1{(z-7/8)^3} - \frac1{(z-1/8)^3} \right ]$$
$f$ has poles at $z_1=1/8$ and $z_2=7/8$.  At $z_1$, the residue is
$$\frac1{2 \cdot 8^3} \frac1{2!} \left [ \frac{d^2}{dz^2} \cot{\pi z} \right ]_{z=1/8} = \frac1{2 \cdot 8^3}  \frac1{2!} (2 \pi ^2) \cot \left(\frac{\pi }{8}\right) \csc ^2\left(\frac{\pi }{8}\right)$$
The calculation for the other pole is similar.  Then our sum is
$$\frac1{2 \cdot 8^3} \frac1{2!} \pi (2 \pi ^2) \left [\cot \left(\frac{\pi }{8}\right) \csc ^2\left(\frac{\pi }{8}\right)-\cot \left(\frac{7 \pi }{8}\right) \csc ^2\left(\frac{7 \pi }{8}\right) \right ] = \frac{\pi^3}{\cdot 8^3} \cot \left(\frac{\pi }{8}\right) \csc ^2\left(\frac{\pi }{8} \right )$$
Simplifying...
$$\cot \left(\frac{\pi }{8}\right) = 1+\sqrt{2}$$
$$\csc ^2\left(\frac{\pi }{8} \right ) = \frac{2}{1-\frac{\sqrt{2}}{2}} = \frac{4}{2-\sqrt{2}} = 2 \left (2+\sqrt{2} \right)$$
Thus:

$$\sum_{n=1}^{\infty} \left [\frac1{(8 n-1)^3} - \frac1{(8 n-7)^3} \right ] = \frac{\pi^3}{256} \left (4+3 \sqrt{2} \right )$$

A: For $b \in \{0,\ldots,7\}$, consider the function $G(b) := \sum_{n \geq 1} \frac{e^{-2\pi i nb/8}}{n^3}$. Observe that
\begin{equation*}
G(b) = \sum_{0 \leq a \leq 7} e^{-2\pi i ab/8} \sum_{n \geq 1} \frac{1}{(8n-a)^3}.
\end{equation*}
Now, we know that if $a,a' \in \{0,\ldots,7\}$ then
\begin{equation*}
\frac{1}{8}\sum_{0 \leq b \leq 7} e^{2\pi i(a-a')b/8} = \begin{cases} 1& \text{ if $a = a'$} \\ 0& \text{ otherwise}. \end{cases}
\end{equation*}
Therefore, we have
\begin{equation*}
\frac{1}{8} \sum_{0 \leq b \leq 7}e^{-2\pi i(7b)/8} G(b) = \sum_{n \geq 1} \frac{1}{(8n-7)^3},
\end{equation*}
and thus your sum in question is precisely 
\begin{equation*}
\frac{1}{8}\sum_{0 \leq b \leq 7} \left(e^{-2\pi i(7b)/8} -e^{-2\pi ib/8}\right)G(b).
\end{equation*}
Now, observe that $G(b) = \sum_{n \geq 1} a_n e^{-2\pi int}$, where $t = \frac{b}{8}$ and $a_n = \frac{1}{n^3}$. This looks like a Fourier series. Can you determine which function it represents? If so then you would be nearly done.
A: This answer uses the hint using the polygamma function
$$
\psi^{(2)}(z) = -\int_0^1 \frac{t^{z-1}}{1-t}\ln^2t dt.
$$
First, using the expansion of $(1-t)^{-1}$, the polygamma function $\psi^{(2)}(x)$ can be written as
\begin{align}
\psi^{(2)}(z) &= -\sum_{n=0}^\infty \int_0^1 t^{n+z-1}\ln^2 tdt \cr
 &= -\sum_{n=0}^\infty \int_0^\infty s^2 e^{-(n+z)s}ds
\qquad \qquad \qquad (t=e^{-s}) \cr
&= -2\sum_{n=0}^\infty \frac{1}{(n+z)^3}.
\end{align}
Therefore, 
$$
\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right) = \frac{1}{1024}\left(\psi^{(2)}\left(\frac{7}{8}\right)
-\psi^{(2)}\left(\frac{1}{8}\right)\right).
$$
Using the reflection relation
$$
\psi^{(2)}(1-z)-\psi^{(2)}(z) = \pi\frac{d^2}{dz^2}\cot \pi z
$$
with $z=1/8$,  the sum can be written as
$$
\left.\frac{\pi}{1024}\frac{d^2}{dz^2}\cot \pi z\right|_{z=1/8}
=\frac{\pi^3}{512}\left(
\cot\frac{\pi}{8}+\cot^3\frac{\pi}{8}\right).
$$
Finally, using trigonometric identities (half angle),we can get $\cot \pi/8 = 1+\sqrt{2}$, and 
$$
\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)
=\frac{\pi^3}{512}\Big(
1+\sqrt{2}+(1+\sqrt{2})^3
\Big)=\frac{\pi^3}{256}(4+3\sqrt{2}).
$$
