Determining value of infinite sum after computing full Fourier Series I have computed the Full Fourier Series of the function $\phi:[-\pi,\pi] \rightarrow \Bbb{R}$ defined by $\forall x \epsilon[-\pi,\pi], \phi(x)=|\sin(x)|$ to be:
$$ \phi(x) = {2\over\pi}+{1\over\pi} \sum_{n=1}^\infty\left[(-1)^{n+1}{2\over{n^2-1}} - {2\over{n^2-1}}\right]\cos(nx)$$
Now I need to compute both of the following:
$\sum_{n=1}^\infty {1\over{4n^2-1}}$ and $\sum_{n=1}^\infty {(-1)^n\over{4n^2-1}}$
I have calculated $\sum_{n=1}^\infty {1\over{4n^2-1}} = {1\over2}$
I'm looking for suggestions on $\sum_{n=1}^\infty {(-1)^n\over{4n^2-1}}$.  I am very rusty on evaluating infinite sums and am not sure how to take the fourier series I calculated and use it to find the sum of the alternating series given.
 A: Expand the term $\frac{1}{4n^2-1}$ using partial fractions:
$$\frac{1}{4n^2-1}=\frac{1}{(2n-1)(2n+1)}=\frac12\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right).$$
This expansion can be used to split up the alternating series $\sum_{n=1}^\infty {(-1)^n\over{4n^2-1}}$ into a sum of two simpler alternating series. In fact, these two series are found to be different tails of the same series. Shifting indices and adding appropriate finite initial terms as needed, the expression can be simplified as follows:
$$\sum_{n=1}^\infty {(-1)^n\over{4n^2-1}}=\frac12\sum_{n=1}^\infty\left(\frac{(-1)^n}{(2n-1)}-\frac{(-1)^n}{(2n+1)}\right)\\
=\frac12\sum_{n=1}^\infty\frac{(-1)^n}{(2n-1)}-\frac12\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)}\\
=\frac12\sum_{n'=0}^\infty\frac{(-1)^{n'+1}}{(2n'+1)}-\frac12\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)}\\
=-\frac12\sum_{n'=0}^\infty\frac{(-1)^{n'}}{(2n'+1)}-\frac12\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)}\\
=\frac12-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)}\\
=\frac12-\frac{\pi}{4}.$$
Note*: The final line uses a well-known series called Leibniz's formula for $\pi$ (wiki). 
