Calculate the values of the serieses

Question

Given this function: $f(x)=|\sin x|$, I was supposed to find its cosinuses series, which came out $$\frac{2}{\pi}+\sum_{k=1}^\infty\frac{4}{\pi\left(1-4k^2\right)}\cos(2kx)\,.$$ After that, I was asked to find the values of these two serieses: $$\sum_{n=1}^\infty\frac{1}{4n^2-1}$$ $$\sum_{n=1}^\infty\frac{n^2}{\left(4n^2-1\right)^2}$$ Using Parseval Theorem, I got to this: $$\sum_{k=1}^\infty\frac{1}{\left(4k^2-1\right)^2}=\frac{3\pi^2}{16}-\frac{1}{2}$$ but I don’t know what to do next…

Answer 1

Since $\;\displaystyle 0=f(0)=\frac{2}{\pi}+\sum_{k=1}^\infty\frac{4}{\pi\left(1-4k^2\right)}\,,\;$ it follows that

$\displaystyle\sum_{k=1}^\infty\frac{4}{\pi\left(4k^2-1\right)}=\frac2\pi\;\;,$

$\displaystyle\sum_{k=1}^\infty\frac1{4k^2-1}=\frac12\;.$

Hence,

$\displaystyle\sum_{n=1}^\infty\frac1{4n^2-1}=\frac12\;.$

Moreover, by using Parseval Theorem, we can get that :

$\displaystyle\sum_{k=1}^\infty\frac{1}{\left(4k^2-1\right)^2}=\frac{\pi^2}{16}-\frac{1}{2}\;.$

Consequently ,

$\displaystyle\sum_{n=1}^\infty\frac{n^2}{\left(4n^2-1\right)^2}= \frac14\left[\sum_{n=1}^\infty\frac{4n^2-1}{\left(4n^2-1\right)^2}+\sum_{n=1}^\infty\frac1{\left(4n^2-1\right)^2}\right]=$

$\displaystyle= \frac14\left[\sum_{n=1}^\infty\frac1{4n^2-1}+\sum_{n=1}^\infty\frac1{\left(4n^2-1\right)^2}\right]=$

$=\dfrac14\left(\dfrac12+\dfrac{\pi^2}{16}-\dfrac12\right)=\dfrac{\pi^2}{64}\;.$