Interesting proofs of this sum So I was asked to solve the following sum using Fourier Series:
$$\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}^{(1)}$$
I decided to try splitting it up into two sums (for even and odd $n$) in the hopes that it would telescope and be easier to evaluate. It did not, but I was able to salvage my work by employing the arctangent Taylor Series. The purpose of this post is twofold: to post a rather unusual approach and to ask others how they might do this problem more easily/differently.
First I noted
$$\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}=\sum_{n\in\mathbb N}\frac{1}{16n^2-1}^{(2)}-\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}^{(3)}$$
Using partial fractions of the first sum on the RHS:
$$(2)\;\sum_{n\in\mathbb N}\frac{1}{16n^2-1}=\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)$$
Using partial fractions on the second sum on the RHS:
$$(3)\;\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}=\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-3}-\frac{1}{4n-1}\right)$$
$$=-\frac{1}{2}+\frac{1}{2}\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)$$
Combining results for the even and odd sums from the RHS:
$$\sum_{n\in\mathbb N}\frac{1}{16n^2-1}-\sum_{n\in\mathbb N}\frac{1}{4(2n-1)^2-1}=-\frac{1}{2}+\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)^{(4)}$$
Then I saw that the rightmost sum was the following:
$$\sum_{n\in\mathbb N}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)=\sum_{n=2}^\infty\frac{(-1)^{n}}{2n-1}^{(5)}$$
Note the Taylor series for arctangent:
$$\text{atan}(x)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{2n-1}$$
Letting $x=1$:
$$\text{atan}(1)=\frac{\pi}{4}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n-1}$$
Accounting for index and sign difference, we see that
$$(5)\;\sum_{n=2}^\infty\frac{(-1)^{n}}{2n-1}=1-\frac{\pi}{4}$$
And finally, from the $-1/2$ term in $(4)$, we see that
$$(1)\;\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}=\frac{2-\pi}{4}$$
 A: I would go about it in a similar way like you do, just shorter, as follows:
By partial fractions, 
$$
\frac{1}{4n^2-1} = \frac12 \left(\frac{1}{2n-1} - \frac{1}{2n+1} \right)
$$
So 
$$
\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}^{(1)} = \frac12 \sum_{n\in\Bbb N} \left(\frac{(-1)^n}{2n-1} - \frac{(-1)^n}{2(n+1)-1} \right)
$$
and by shifting the sum index
$$
\sum_{n\in\Bbb N}\frac{(-1)^n}{4n^2-1}^{(1)} = \frac12  \left(
\sum_{n=1}^\infty \frac{(-1)^n}{2n-1} + \sum_{n=2}^\infty \frac{(-1)^n}{2n-1} \right)\\
= \frac12  \left( 1 -
2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1} \right) = \frac{2-\pi}{4}
$$
where in the last line the $\arctan$ result was used.
A: Here is my attempt to follow the instruction to use Fourier series.  First, as you did, I rewrite the sum as
$$\sum_{n=1}^\infty\,\frac{(-1)^n}{4n^2-1}=\sum_{n=1}^\infty\,\frac{(-1)^n}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}+\sum_{n=1}^\infty\,\frac{(-1)^n}{2n-1}\,.\tag{*}$$
Let $f:\mathbb{R}\to\mathbb{R}$ be the periodic function with period $2\pi$ defined by
$$f(x):=\begin{cases}+1&\text{if }x\in(0,+\pi)\,,\\0&\text{if }x\in\{-\pi,0,+\pi\}\,,\\-1&\text{if }x\in(-\pi,0)\,.\end{cases}$$
Then, as $f$ is an odd function, $f$ has a Fourier sine-series:
$$f(x)=\sum_{k=1}^\infty\,b_k\,\sin(kx)$$
for some real numbers $b_1,b_2,b_3,\ldots$.  We see that
$$\begin{align}b_k&=\frac{1}{\pi}\,\int_{-\pi}^{+\pi}\,\sin(kx)\,f(x)\,\text{d}x=\frac{2}{\pi}\,\int_0^\pi\,\sin(kx)\,\text{d}x
\\&=\frac{2}{\pi\,k}\,\Big.\big(-\cos(kx)\big)\Big|_{x=0}^{x=\pi}=\frac{2\,\big(1-(-1)^k\big)}{\pi\,k}
\end{align}$$
for all $k=1,2,3,\ldots$.
Consequently,
$$f(x)=\sum_{k=1}^\infty\,\frac{2\,\big(1-(-1)^k\big)}{\pi\,k}\,\sin(kx)=\frac{4}{\pi}\,\sum_{n=1}^\infty\,\frac{\sin\big((2n-1)x\big)}{2n-1}\,.$$
Because $f\left(\dfrac{\pi}{4}\right)=1$, we get
$$1=\frac{4}{\pi}\,\sum_{n=1}^\infty\,\frac{\sin\left(\frac{(2n-1)\,\pi}{4}\right)}{2n-1}=\frac{4}{\pi}\,\sum_{n=1}^\infty\,\frac{(-1)^{n-1}}{2n-1}\,.$$
This shows that
$$\sum_{n=1}^\infty\,\frac{(-1)^n}{2n-1}=-\frac{\pi}{4}\,.$$
By (*), we conclude that
$$\sum_{n=1}^\infty\,\frac{(-1)^n}{4n^2-1}=\frac{1}{2}-\frac{\pi}{4}=\frac{2-\pi}{4}\approx -0.28539816\,.$$
A: Consider a $2\pi$-periodic function $f:\Bbb{R}\to\Bbb{C}$ whose values on $(-\pi,\pi]$ are given by $f(x)=\operatorname{sign}x\sin\alpha x$.  Then, $f$ is even so it has a Fourier series of the form
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx,$$
where $a_n=\frac1\pi\int_{-\pi}^\pi f(x)\cos nx\ dx=\frac{2}{\pi}\int_0^\pi f(x)\cos nx\ dx$ (recalling that $f$ is even).  Because $2\sin\alpha x\cos nx=\sin(n+\alpha)x-\sin(n-\alpha)x$, we conclude that
$$a_n=\begin{cases}\frac{1-(-1)^n\cos\alpha \pi}{\pi}\left(\frac{1}{n+\alpha}-\frac{1}{n-\alpha}\right)&\text{if }n\neq \pm\alpha,\\ 0&\text{if }n=\pm\alpha.\end{cases}$$
Take $\alpha=1/2$, so that
$$a_n=\frac{1-(-1)^n\cos\frac{\pi}{2}}{\pi}\left(\frac{1}{n+1/2}-\frac{1}{n-1/2}\right)=-\frac{4}{\pi(4n^2-1)}.$$
That is
$$\operatorname{sign}x\sin\frac{x}{2}=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx=\frac2\pi-\frac4\pi\sum_{n=1}^\infty\frac{1}{4n^2-1}\cos nx$$
for $-\pi< x\le \pi$.  Plugging in $x=\pi$ (note that $f$ is continuous at $x=\pi$, so we don't have to worry about jumps over there), we have
$$1=\operatorname{sign}{\pi}\sin\frac{\pi}{2}=\frac2\pi-\frac4\pi\sum_{n=1}^\infty\frac{\cos n\pi}{4n^2-1}=\frac2\pi-\frac4\pi\sum_{n=1}^\infty\frac{(-1)^n}{4n^2-1}.$$
So $\sum_{n=1}^\infty\frac{(-1)^n}{4n^2-1}=\frac{2-\pi}{4}$.

Similarly, consider a $2\pi$-periodic function $g:\Bbb{R}\to\Bbb{C}$ whose values on $(-\pi,\pi]$ are given by $g(x)=\operatorname{sign}x\cos\alpha x$.  Then, $g$ is odd so it has a Fourier series of the form
$$g(x)=\sum_{n=1}^\infty b_n\sin nx,$$
where $b_n=\frac1\pi\int_{-\pi}^\pi g(x)\sin nx\ dx=\frac{2}{\pi}\int_0^\pi g(x)\sin nx\ dx$ (recalling that $g$ is odd).  Because $2\cos\alpha x\sin nx=\sin(n+\alpha)x+\sin(n-\alpha)x$, we conclude that
$$b_n=\begin{cases}\frac{1-(-1)^n\cos\alpha \pi}{\pi}\left(\frac{1}{n+\alpha}+\frac{1}{n-\alpha}\right)&\text{if }n\neq \pm\alpha,\\ 0&\text{if }n=\pm\alpha.\end{cases}$$
