# Show that $\frac3{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3}=-\frac{105}{64}\zeta(3)$

Consider the infinite sum

$$\frac34\sum_{n=1}^{\infty}\left[\frac{(-1)^n}{(2n)^3}-\frac2{n^3}\right]=\frac3{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3}$$

I have come across this sum while evaluating the integral

$$\int_{0}^{\pi/4}\frac{x^3}{\sin^2x}dx=\frac{3\pi}{4}G-\frac{\pi^3}{64}+\frac{3\pi^2}{32}\log2-\frac{105}{64}\zeta(3)$$

where $$G$$ denotes Catalan's Constant and $$\zeta(z)$$ the Riemann Zeta Function. To do so I followed this approach (Solution to Problem $$18$$ given by ysharifi)

Integration by parts with $$x^3=u$$ and $$\frac{dx}{\sin^2 x}=dv$$ reduces the problem to $$\int_0^{\pi/4}x^2\cot x~dx$$. Again, integration by parts with $$x^2=u$$ and $$\cot x~dx=dv$$ reduces the problem to $$\int_0^{\pi/4}x\ln \sin x~dx$$ and this one can by easily found using the identity $$\ln \sin x =-\ln 2-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}$$ which holds for $$x\in(0,\pi)$$

The sum I failed to express in terms of the Riemann Zeta Function occured within the last step of calculation. To put in a nutshell I do not know how to show

$$\frac3{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3}=-\frac{105}{64}\sum_{n=1}^{\infty}\frac1{n^3}=-\frac{105}{64}\zeta(3)$$

Hence this is the last step of evaluating the integral I would be glad if someone could explain to me how to show the equalitiy of these two sums.

• $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}=\sum_{n=1}^{\infty}\frac{1}{n^3}-2\sum_{n=1}^{\infty}\frac{1}{(2n)^3}$. – metamorphy Sep 27 '18 at 12:29
• $$\frac34\sum_{n=1}^{\infty}\left[\frac{(-1)^n}{(2n)^3}-\frac2{n^3}\right]\color{red}{\ne}\frac3{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3}$$ – Hazem Orabi Sep 27 '18 at 13:54
\begin{align} \color{red}{S} & =+\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3} =-\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}-\frac{3}{16}\sum_{n=1}^{\infty}\frac{8}{n^3} \\[2mm] & =-\frac{3}{16}\eta(3)-\frac{3}{2}\zeta(3) =\left(-\frac{3}{16}\left(1-2^{1-3}\right)-\frac{3}{2}\right)\zeta(3) =\color{red}{-\frac{105}{64}\zeta(3)} \end{align}