Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$

Question

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.

Answer 1

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:

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

Answer 2

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.

Answer 3

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 $$ \sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right) =\frac{\pi^3}{512}\Big( 1+\sqrt{2}+(1+\sqrt{2})^3 \Big)=\frac{\pi^3}{256}(4+3\sqrt{2}). $$