Csdrhrt

So I was asked to solve the following sum using Fourier Series:

$$sum_{ninBbb N}frac{(-1)^n}{4n^2-1}^{(1)}$$

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

$$sum_{ninBbb N}frac{(-1)^n}{4n^2-1}=sum_{ninmathbb N}frac{1}{16n^2-1}^{(2)}-sum_{ninmathbb N}frac{1}{4(2n-1)^2-1}^{(3)}$$

Using partial fractions of the first sum on the RHS:

$$(2);sum_{ninmathbb N}frac{1}{16n^2-1}=frac{1}{2}sum_{ninmathbb N}left(frac{1}{4n-1}-frac{1}{4n+1}right)$$

Using partial fractions on the second sum on the RHS:

$$(3);sum_{ninmathbb N}frac{1}{4(2n-1)^2-1}=frac{1}{2}sum_{ninmathbb N}left(frac{1}{4n-3}-frac{1}{4n-1}right)=-frac{1}{2}+frac{1}{2}sum_{ninmathbb N}left(frac{1}{4n-1}-frac{1}{4n+1}right)$$

Combining results for the even and odd sums from the RHS:

$$sum_{ninmathbb N}frac{1}{16n^2-1}-sum_{ninmathbb N}frac{1}{4(2n-1)^2-1}=-frac{1}{2}+sum_{ninmathbb N}left(frac{1}{4n-1}-frac{1}{4n+1}right)^{(4)}$$

Then I saw that the rightmost sum was the following:

$$sum_{ninmathbb N}left(frac{1}{4n-1}-frac{1}{4n+1}right)=sum_{n=2}^inftyfrac{(-1)^{n}}{2n-1}^{(5)}$$

Note the Taylor series for arctangent:

$$text{atan}(x)=sum_{n=1}^inftyfrac{(-1)^{n-1}x^n}{2n-1}$$

Letting $x=1$:

$$text{atan}(1)=frac{pi}{4}=sum_{n=1}^inftyfrac{(-1)^{n+1}}{2n-1}$$

Accounting for index and sign difference, we see that

$$(5);sum_{n=2}^inftyfrac{(-1)^{n}}{2n-1}=1-frac{pi}{4}$$

And finally, from the $-1/2$ term in $(4)$, we see that

$$(1);sum_{ninBbb N}frac{(-1)^n}{4n^2-1}=frac{2-pi}{4}$$

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
$$
sum_{ninBbb N}frac{(-1)^n}{4n^2-1}^{(1)} = frac12 sum_{ninBbb N} left(frac{(-1)^n}{2n-1} - frac{(-1)^n}{2(n+1)-1} right)
$$
and by shifting the sum index
$$
sum_{ninBbb N}frac{(-1)^n}{4n^2-1}^{(1)} = frac12 left(
sum_{n=1}^infty frac{(-1)^n}{2n-1} + sum_{n=2}^infty frac{(-1)^n}{2n-1} right)\
= frac12 left( 1 -
2 sum_{n=1}^infty frac{(-1)^{n+1}}{2n-1} right) = frac{2-pi}{4}
$$
where in the last line the $arctan$ result was used.

Consider a $2pi$-periodic function $f:Bbb{R}toBbb{C}$ whose values on $(-pi,pi]$ are given by $f(x)=operatorname{sign}xsinalpha x$. Then, $f$ is even so it has a Fourier series of the form
$$f(x)=frac{a_0}{2}+sum_{n=1}^infty a_ncos nx,$$
where $a_n=frac1piint_{-pi}^pi f(x)cos nx dx=frac{2}{pi}int_0^pi f(x)cos nx dx$ (recalling that $f$ is even). Because $2sinalpha xcos nx=sin(n+alpha)x-sin(n-alpha)x$, we conclude that
$$a_n=begin{cases}frac{1-(-1)^ncosalpha pi}{pi}left(frac{1}{n+alpha}-frac{1}{n-alpha}right)&text{if }nneq pmalpha,\ 0&text{if }n=pmalpha.end{cases}$$

Take $alpha=1/2$, so that
$$a_n=frac{1-(-1)^ncosfrac{pi}{2}}{pi}left(frac{1}{n+1/2}-frac{1}{n-1/2}right)=-frac{4}{pi(4n^2-1)}.$$
That is
$$operatorname{sign}xsinfrac{x}{2}=frac{a_0}{2}+sum_{n=1}^infty a_ncos nx=frac2pi-frac4pisum_{n=1}^inftyfrac{1}{4n^2-1}cos nx$$
for $-pi< xle pi$. 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
$$1=operatorname{sign}{pi}sinfrac{pi}{2}=frac2pi-frac4pisum_{n=1}^inftyfrac{cos npi}{4n^2-1}=frac2pi-frac4pisum_{n=1}^inftyfrac{(-1)^n}{4n^2-1}.$$
So $sum_{n=1}^inftyfrac{(-1)^n}{4n^2-1}=frac{2-pi}{4}$.

Similarly, consider a $2pi$-periodic function $g:Bbb{R}toBbb{C}$ whose values on $(-pi,pi]$ are given by $g(x)=operatorname{sign}xcosalpha x$. Then, $g$ is odd so it has a Fourier series of the form
$$g(x)=sum_{n=1}^infty b_nsin nx,$$
where $b_n=frac1piint_{-pi}^pi g(x)sin nx dx=frac{2}{pi}int_0^pi g(x)sin nx dx$ (recalling that $g$ is odd). Because $2cosalpha xsin nx=sin(n+alpha)x+sin(n-alpha)x$, we conclude that
$$b_n=begin{cases}frac{1-(-1)^ncosalpha pi}{pi}left(frac{1}{n+alpha}+frac{1}{n-alpha}right)&text{if }nneq pmalpha,\ 0&text{if }n=pmalpha.end{cases}$$

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 $f:mathbb{R}tomathbb{R}$ be the periodic function with period $2pi$ defined by
$$f(x):=begin{cases}+1&text{if }xin(0,+pi),,\0&text{if }xin{-pi,0,+pi},,\-1&text{if }xin(-pi,0),.end{cases}$$
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
$$begin{align}b_k&=frac{1}{pi},int_{-pi}^{+pi},sin(kx),f(x),text{d}x=frac{2}{pi},int_0^pi,sin(kx),text{d}x
\&=frac{2}{pi,k},Big.big(-cos(kx)big)Big|_{x=0}^{x=pi}=frac{2,big(1-(-1)^kbig)}{pi,k}
end{align}$$
for all $k=1,2,3,ldots$.
Consequently,
$$f(x)=sum_{k=1}^infty,frac{2,big(1-(-1)^kbig)}{pi,k},sin(kx)=frac{4}{pi},sum_{n=1}^infty,frac{sinbig((2n-1)xbig)}{2n-1},.$$
Because $fleft(dfrac{pi}{4}right)=1$, we get
$$1=frac{4}{pi},sum_{n=1}^infty,frac{sinleft(frac{(2n-1),pi}{4}right)}{2n-1}=frac{4}{pi},sum_{n=1}^infty,frac{(-1)^{n-1}}{2n-1},.$$
This shows that
$$sum_{n=1}^infty,frac{(-1)^n}{2n-1}=-frac{pi}{4},.$$
By (*), we conclude that
$$sum_{n=1}^infty,frac{(-1)^n}{4n^2-1}=frac{1}{2}-frac{pi}{4}=frac{2-pi}{4}approx -0.28539816,.$$