$\sum_{n=1}^{\infty}\frac{(-1)^n}{16n^2-1}$

How can I calculate $$\sum_{n=1}^{\infty}\frac{(-1)^n}{16n^2-1}$$ using the Fourier series?

Solution 1:

\begin{align} |\sin(x)|&=\frac2{\pi}-\frac4{\pi}\sum_{n=1}^\infty\frac{\cos(2nx)}{4n^2-1}\\\\ \therefore\left|\sin\left(\frac\pi 4\right)\right|&=\frac2{\pi}-\frac4{\pi}\sum_{n=1}^\infty\frac{\cos\left(\frac{n\pi} 2\right)}{4n^2-1}\\ &=\frac2{\pi}-\frac4{\pi}\sum_{n=even}^\infty\frac{(-1)^{n/2}}{4n^2-1}\\ (n\rightarrow 2m)&=\frac2{\pi}-\frac4{\pi}\sum_{m=1}^\infty\frac{(-1)^m}{16m^2-1}\\\\ \sum_{n=1}^\infty\frac{(-1)^n}{16n^2-1}&=\frac{\frac2{\pi} - \frac{\sqrt2}2 }{\frac4\pi} \end{align}