Sum of a speciale series

Question

I would like to know if what i have found is correct or not

what is the value of this series

$$\sum_{n=2,n\;\text{even} }^{\infty}\dfrac{\sin^{2}(nt)}{n^2(n^2-1)}$$

where $t \in [0 , 2\pi]$

Thanks in adavance

Answer 1

As $n\ge 2$ even integer we can write $n=2k$ where $k\ge1$:

$\sum\limits_{k=1 }^{\infty}\dfrac{\sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=\dfrac{1}{2}\sum\limits_{k=1 }^{\infty}\dfrac{1-\cos(4kt)}{(2k)^2-1}-\dfrac{1}{2}\sum\limits_{k=1 }^{\infty}\dfrac{1-\cos(4kt)}{(2k)^2}\tag1$

Regrouping (1):

$\dfrac{1}{2}\sum\limits_{k=1 }^{\infty}\dfrac{1}{(2k)^2-1}-\dfrac{1}{8}\sum\limits_{k=1 }^{\infty}\dfrac{1}{k^2}-\dfrac{1}{2}\sum\limits_{k=1 }^{\infty}\dfrac{\cos(4kt)}{(2k)^2-1}+\dfrac{1}{8}\sum\limits_{k=1 }^{\infty}\dfrac{\cos(4kt)}{k^2}\tag2$

Let's calculate the sums of (2) one after the other:

• First one is telescoping $S_1=\dfrac{1}{2}\sum\limits_{k=1 }^{\infty}\dfrac{1}{(2k)^2-1}=\dfrac{1}{4}\sum\limits_{k=1 }^{\infty}\Big(\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\big)=\dfrac{1}{4}\tag{2.1}$

• The second one is Riemann zeta function:

$S_2=-\dfrac{1}{8}\sum\limits_{k=1 }^{\infty}\dfrac{1}{k^2}=\dfrac{-\zeta(2)}{8}=-\dfrac{\pi^2}{48}\tag{2.2}$

• At the third one we have to realize that the sum is proportional to the Fourier series of

$| \sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok; $|\sin(2t)|=\dfrac{2}{\pi}-\dfrac{4}{\pi}\sum\limits_{k=1 }^{\infty}\dfrac{\cos(4kt)}{(2k)^2-1}$

So $S_3=-\dfrac{1}{4}+\dfrac{\pi}{8}|\sin(2t)|\tag{2.3}$

• The last one can be expressed with dilogarithm function:

As $\cos(4kt)=\dfrac{e^{4ikt}+e^{-4ikt}}{2}$

$S_4=-\dfrac{1}{16}\Big(Li_2(e^{4it})+Li_2(e^{-4it})\Big)=-\dfrac{1}{8}Ci_2(4t)\tag{2.4}$

where $Ci_2$ is the real part of Clausen function of order 2.

Finally the result:

$\sum_{n=2,n\;\text{even} }^{\infty}\dfrac{\sin^{2}(nt)}{n^2(n^2-1)}=\dfrac{\pi}{8}|\sin(2t)|-\dfrac{\pi^2}{48}-\dfrac{1}{8}Ci_2(4t)\tag3$