Consider the infinite sum


I have come across this sum while evaluating the integral


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


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.

Thanks in advance!

  • 1
    $\begingroup$ $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}=\sum_{n=1}^{\infty}\frac{1}{n^3}-2\sum_{n=1}^{\infty}\frac{1}{(2n)^3}$. $\endgroup$ – metamorphy Sep 27 '18 at 12:29
  • $\begingroup$ @metamorphy First of all thank you for your quick response- I guess I understand your attempt but I am not sure whether one can do this so easily or not. Hence we are talking about infinite sums; and I have had my problems with simply splitting them up. Nevertheless I think this possible because the sum converges absolutely (?) but I am not sure about this. Could you maybe go further by explaining your whole idea in more than just one line :) $\endgroup$ – mrtaurho Sep 27 '18 at 12:35
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    $\begingroup$ @mrtaurho Exactly. Absolute convergence allows us to rearrange the terms arbitrarily. $\endgroup$ – Szeto Sep 27 '18 at 12:37
  • $\begingroup$ @Szeto Ah okay. I guess from hereon it is kind of trivial. However could someone put this all together within an answer refering to the sum I gave? :) $\endgroup$ – mrtaurho Sep 27 '18 at 12:39
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    $\begingroup$ $$ \frac34\sum_{n=1}^{\infty}\left[\frac{(-1)^n}{(2n)^3}-\frac2{n^3}\right]\color{red}{\ne}\frac3{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3} $$ $\endgroup$ – Hazem Orabi Sep 27 '18 at 13:54

$$ \begin{align} \color{red}{S} & =+\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3} =-\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}-\frac{3}{16}\sum_{n=1}^{\infty}\frac{8}{n^3} \\[2mm] & =-\frac{3}{16}\eta(3)-\frac{3}{2}\zeta(3) =\left(-\frac{3}{16}\left(1-2^{1-3}\right)-\frac{3}{2}\right)\zeta(3) =\color{red}{-\frac{105}{64}\zeta(3)} \end{align} $$


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