I have been given the task of showing that $\sum_{-\infty}^{\infty} \frac{(-1)^{n-1}}{n^2-\frac{1}{4}} = 2\pi$ and quite frankly I am stuck. So far I have shown that $|c_n|^2 = \sum_{-\infty}^{\infty} \frac{1}{(n^2-\frac{1}{4})^2} = 2\pi^2$ by using Parseval's Identity with the function $f(x) = |\cos(\frac{x}{2})|$. I hope someone here is able to help in advance thanks for your help.
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This thread may help (consider the Fourier series for $,z:=\dfrac 12$). – Raymond Manzoni Oct 26 '18 at 12:03
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Or you can apply the same method to $\sum_n \frac{1}{((2n)^2-1/4)^2}$ and subtract the two series – reuns Oct 26 '18 at 12:55