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
$$f(x)=\sum_{n=1}^{\infty}b_n\sin\left(\frac{\pi n x}{l}\right),\tag{1}$$
where
$$b_n=\frac{2}{l}\int\limits_{0}^{l}f(x)\sin\left(\frac{\pi n x}{l}\right).\tag{2}$$
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}\int\limits_{0}^{\pi/2}\cos(x)\sin(2nx)dx = \frac{4(\sin(\pi n)-2n)}{\pi(1-4n^2)}\tag{3}.
\end{align}
This gives
$$\cos(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^2-1}\sin(2nx), \quad \forall \ n\in\mathbb{N}\tag{4}$$
2 questions remain:
- 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?
- How do I calculate the desired sum?
Best Answer
As posted by the OP, two steps remain:
$$ b_n=\frac{4}{\pi}\int\limits_{0}^{\pi/2}\cos(x)\sin(2nx)dx \\ = \frac{1}{i \pi}\int\limits_{0}^{\pi/2}e^{ix(1+2n)}- e^{ix(1-2n)}+e^{-ix(1-2n)}-e^{-ix(1+2n)}dx \\ = \frac{1}{i \pi} \left(\frac{e^{ix(1+2n)}}{i(1+2n)}|_{0}^{\pi/2} - \frac{e^{ix(1-2n)}}{i(1-2n)}|_{0}^{\pi/2} + \frac{e^{-ix(1-2n)}}{-i(1-2n)}|_{0}^{\pi/2} -\frac{e^{-ix(1+2n)}}{-i(1+2n)}|_{0}^{\pi/2} \right)\\ = \frac{1}{i \pi} \left(\frac{i(-1)^n - 1}{i(1+2n)} - \frac{i(-1)^n - 1}{i(1-2n)} + \frac{-i(-1)^n - 1}{-i(1-2n)} - \frac{-i(-1)^n - 1}{-i(1+2n)} \right)\\ = \frac{1}{- \pi (1 - 4 n^2)} \left( -4n(i(-1)^n - 1) - 4n (-i(-1)^n - 1) \right)\\ = \frac{-8n}{\pi(1-4n^2)} \\ = \frac{4(\sin(\pi n)-2n)}{\pi(1-4n^2)} $$ 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:
$$ \frac{\pi}{4} = \int\limits_{0}^{\pi/2}\cos^2(x) dx = \frac{8}{\pi} \sum_{n=1}^\infty \frac{n}{4n^2 -1} \int\limits_{0}^{\pi/2}\cos(x)\cdot\sin(2nx) dx = \\ \frac{8}{\pi} \sum_{n=1}^\infty \frac{n}{4n^2 -1} \cdot \frac{2 n}{4n^2 -1}\\ = \frac{16}{\pi} \sum_{n=1}^\infty \frac{n^2}{(4n^2 -1)^2} $$ so finally
$$ \sum_{n=1}^\infty \frac{n^2}{(4n^2 -1)^2} = \frac{\pi^2}{64} $$