Cfrgtkky

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.