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Expand $cos(x)$ in sinus-series in the interval $(0,pi/2)$ and use
the result to calculate $$sum_{n=1}^{infty}frac{n^2}{(4n^2-1)^2}.$$

Since we only need the odd terms in the Fourier expansion, we know that $a_n=0$, so we only need $b_n$ in order to find the sine expansion.

In my book, there is a chapter on "Fourier series on intervals", and there they state that:

If $f$ is a piecewise smooth function on $[0,l]$, then the Fourier sine expansion is given by

$$f(x)=sum_{n=1}^{infty}b_nsinleft(frac{pi n x}{l}right),tag{1}$$

where

$$b_n=frac{2}{l}intlimits_{0}^{l}f(x)sinleft(frac{pi n x}{l}right).tag{2}$$

So we should be able to use only $(1)$ and $(2)$ with $l=pi/2$ to solve this problem. So

begin{align}
b_n=frac{4}{pi}intlimits_{0}^{pi/2}cos(x)sin(2nx)dx = frac{4(sin(pi n)-2n)}{pi(1-4n^2)}tag{3}.
end{align}

This gives

$$cos(x)=frac{8}{pi}sum_{n=1}^{infty}frac{n}{4n^2-1}sin(2nx), quad forall ninmathbb{N}tag{4}$$

2 questions remain:

1. I used wolfram alpha to calculate the integral in $(3)$. What methods do I use to do this? Repeated integration by parts and solve for the integral?


2. How do I calculate the desired sum?

As posted by the OP, two steps remain:

1. Perform the integral:

$$
b_n=frac{4}{pi}intlimits_{0}^{pi/2}cos(x)sin(2nx)dx \
= frac{1}{i pi}intlimits_{0}^{pi/2}e^{ix(1+2n)}- e^{ix(1-2n)}+e^{-ix(1-2n)}-e^{-ix(1+2n)}dx \
= frac{1}{i pi} left(frac{e^{ix(1+2n)}}{i(1+2n)}|_{0}^{pi/2} - frac{e^{ix(1-2n)}}{i(1-2n)}|_{0}^{pi/2} + frac{e^{-ix(1-2n)}}{-i(1-2n)}|_{0}^{pi/2} -frac{e^{-ix(1+2n)}}{-i(1+2n)}|_{0}^{pi/2} right)\
= frac{1}{i pi} left(frac{i(-1)^n - 1}{i(1+2n)} - frac{i(-1)^n - 1}{i(1-2n)} + frac{-i(-1)^n - 1}{-i(1-2n)} - frac{-i(-1)^n - 1}{-i(1+2n)} right)\
= frac{1}{- pi (1 - 4 n^2)} left( -4n(i(-1)^n - 1) - 4n (-i(-1)^n - 1) right)\
= frac{-8n}{pi(1-4n^2)} \
= frac{4(sin(pi n)-2n)}{pi(1-4n^2)}
$$
where the last step only indicates the "desired" solution by WolframAlpha, where anyway $sin(pi n) = 0$ .

2 Calculating the desired sum.

From OP's last result,
$$
cos(x)=frac{8}{pi}sum_{n=1}^{infty}frac{n}{4n^2-1}sin(2nx),
$$
multiply this formula with $cos(x) $ and integrate, using the integral which was just derived:

$$
frac{pi}{4} = intlimits_{0}^{pi/2}cos^2(x) dx = frac{8}{pi} sum_{n=1}^infty frac{n}{4n^2 -1} intlimits_{0}^{pi/2}cos(x)cdotsin(2nx) dx = \
frac{8}{pi} sum_{n=1}^infty frac{n}{4n^2 -1} cdot frac{2 n}{4n^2 -1}\
= frac{16}{pi} sum_{n=1}^infty frac{n^2}{(4n^2 -1)^2}
$$
so finally

$$
sum_{n=1}^infty frac{n^2}{(4n^2 -1)^2} = frac{pi^2}{64}
$$

Consider the function $f(x) = cos(x) cdot {rm{sign}}(x)$ in the interval $(-pi, pi)$. This function is identical to $cos(x)$ in the interval $(0, pi/2)$ so the expansion will be correct in that interval.

You want $f(x) = sum_{m=1}^infty b_m sin(mx)$.

Integrating with $sin(nx)$ gives

$$
intlimits_{-pi}^{pi}sin( n x)sin(nx)dx = pi
$$
$$
intlimits_{-pi}^{pi}sin( m x)sin(nx)dx = 0 quad (mne n)
$$
and

$$
intlimits_{-pi}^{pi}cos(x)cdot {rm{sign}}(x) cdot sin(nx)dx = 2 intlimits_{0}^{pi}cos(x)cdot sin(nx)dx = 2 frac{n (1+cos(pi n))}{n^2 - 1} tag{*}
$$

So you have
$$
cos(x) = frac{2}{pi} sum_{n=1}^infty frac{n (1+cos(pi n))}{n^2 - 1} sin(nx)
$$
where the coefficients are nonzero for even n.

Substituting $n$ with $2 m$ gives
$$
cos(x) = frac{8}{pi} sum_{m=1}^infty frac{m}{4m^2 -1} sin(2 m x)
$$

No use the integral in (*) again: multiply the last formula with $cos(x) $ and integrate:

$$
frac{pi}{2} = intlimits_{0}^{pi}cos^2(x) dx = frac{8}{pi} sum_{m=1}^infty frac{m}{4m^2 -1} intlimits_{0}^{pi}cos(x)cdotsin(2mx) dx = frac{32}{pi} sum_{m=1}^infty frac{m^2}{(4m^2 -1)^2}
$$
so finally

$$
sum_{m=1}^infty frac{m^2}{(4m^2 -1)^2} = frac{pi^2}{64}
$$