<s>[INST] Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$ I am struggling to show that $$\dfrac \pi{2\sqrt3}=1-\dfrac 15+\dfrac 17-\dfrac 1{11}+\dfrac 1{13}-\dfrac 1{17}+\dfrac 1{19}-\cdots$$ by using the Fourier series $$\frac \pi2-\frac x2=\sum_1^\infty \dfrac {\sin(nx)}{n}.$$
Can somebody give me any hint?
 [/INST] The approach we take is to find a discrete and finite Fourier-series of the periodic sequence $f(n)$ in the sum $\sum\frac{f(n)}{n} = \frac{\color{red}{1}}{1} + \frac{\color{red}{0}}{2} + \frac{\color{red}{0}}{3} + \frac{\color{red}{0}}{4} + \frac{\color{red}{-1}}{5} + \frac{\color{red}{0}}{6} \ldots$ we are trying to compute. This will allow us to use this (finite) Fourier series in combination with the Fourier series in the question to evaluate the sum.

Consider the function $f:\mathbb{N}\to\mathbb{N}$ given by (which is the Dirichlet character modulo $6$ hinted to by Paul in the comments)
$$f(n) = \left\{\matrix{0 & n\equiv 0\mod 6\\1 & n\equiv 1\mod 6\\0 & n\equiv 2\mod 6\\0 & n\equiv 3\mod 6\\0 & n\equiv 4\mod 6\\-1 & n\equiv 5\mod 6}\right.$$
This function is constructed such that your series can be written as
$$1 - \frac{1}{5} + \frac{1}{7} - \frac{1}{11} + \frac{1}{13} - \ldots = \sum_{n=1}^\infty \frac{f(n)}{n}$$
Since $f$ is periodic with period $6$ it has a Fourier series
$$f(n) = \sum_{k=0}^5c_k \sin\left(\frac{2\pi k}{6}n\right)$$
where the coefficients are given by
$$c_k = \frac{1}{6}\sum_{j=0}^5f(j)\sin\left(\frac{2\pi k}{6}j\right) = \frac{\sin\left(\frac{\pi k}{3}\right)-\sin\left(\frac{5\pi k}{3}\right)}{6}$$
The sum you are after can therefore be written as
$$\sum_{n=1}^\infty\frac{f(n)}{n} = \sum_{k=0}^5c_k \color{red}{\sum_{n=1}^\infty\frac{\sin\left(\frac{2\pi k}{6} n\right)}{n}} = \sum_{k=0}^5c_k\color{red}{\left(\frac{\pi}{2} - \frac{\left(\frac{2\pi k}{6}\right)}{2}\right)}$$
where we have used the Fourier series in the question to evaluate the infinite sums. This gives the desired result
$$\sum_{n=1}^\infty \frac{f(n)}{n} = \sum_{k=0}^5\frac{\sin\left(\frac{\pi k}{3}\right)-\sin\left(\frac{5\pi k}{3}\right)}{6} \frac{(3-k)\pi}{6} = \frac{\pi}{2\sqrt{3}}$$

The method we used above generalises to computing the sum of any series $\sum \frac{f(n)}{n}$, where $f(n)$ is a periodic function (with integer period).
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