Determine the sum $\sum_{n=1}^\infty \left( \frac{1}{(4n)^2-1}-\frac{1}{(4n+2)^2+1}\right)$ using the Fourier series of $|\cos x|$ I need to determine the sum $$\sum_{n=1}^\infty  \left( \frac{1}{(4n)^2-1}-\frac{1}{(4n+2)^2+1}\right)$$ using the Fourier series of $\lvert \cos x\rvert$ on the interval $ [-\pi,\pi]$.
I have already calculated Fourier series and I get this:
$$\lvert \cos x\rvert= {\frac2\pi}+\sum_{n=2}^\infty{\frac4\pi}\frac{\cos(n\frac\pi2)}{1-n^2}\cos(nx)$$ 
I do not know how to manipulate the Fourier series to get that specific sum.
I tried this but did not get me anywhere. 
$$\sum_{n=2}^\infty{\frac4\pi}\frac{{\frac12}\cos(n\frac\pi2-nx)-{\frac12}\cos(n\frac\pi2+nx)}{1-n^2}$$
$$-\sum_{n=2}^\infty{\frac4\pi}\frac{\cos(n\frac\pi2-nx)}{2n^2-2}-\frac{\cos(n\frac\pi2+nx)}{2n^2-2}$$
 A: The sum of the first term is
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(4n)^2-1}
&=\frac12\sum_{n=1}^\infty\left(\frac1{4n-1}-\frac1{4n+1}\right)\\
&=\frac12\left[1+\sum_{n\in\mathbb{Z}}\frac1{4n-1}\right]\\
&=\frac12+\frac18\sum_{n\in\mathbb{Z}}\frac1{n-\frac14}\\
&=\frac12-\frac\pi8\cot\left(\frac\pi4\right)\\[6pt]
&=\frac12-\frac\pi8\tag1
\end{align}
$$
The sum of the second term is
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(4n+2)^2+1}
&=\frac i2\sum_{n=1}^\infty\left(\frac1{i-(4n+2)}+\frac1{i+(4n+2)}\right)\\
&=\frac i2\left[\sum_{n\in\mathbb{Z}}\frac1{i+(4n+2)}-\frac1{i+2}-\frac1{i-2}\right]\\
&=\frac i8\sum_{n\in\mathbb{Z}}\frac1{n+\frac{2+i}4}-\frac15\\
&=\frac{\pi i}8\cot\left(\pi\frac{2+i}4\right)-\frac15\\[3pt]
&=\frac\pi8\tanh\left(\frac\pi4\right)-\frac15\tag2
\end{align}
$$
where we have used $(7)$ from this answer (which uses complex analysis) or Theorem $2$ from this answer (which uses Fourier Series).
Subtracting $(2)$ from $(1)$ yields
$$
\sum_{n=1}^\infty\left(\frac1{(4n)^2-1}-\frac1{(4n+2)^2+1}\right)=\frac7{10}-\frac\pi8-\frac\pi8\tanh\left(\frac\pi4\right)\tag3
$$
