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?

This sum may be tackled using the residue theorem.
$$S=\sum_{k=0}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right )  = \frac12 \sum_{k=-\infty}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right )$$
Thus, consider the contour integral
$$\oint_{C_N} dz \frac{2 \pi \, \cot{\pi z}}{36 \left (z+\frac16\right ) \left (z+\frac{5}{6}\right )} $$
where $C_N$ is the square with vertices at $z =\pm (N \pm 1/2)$.  As $N \to \infty$, the integral is zero.  Thus the sum of the residues at the poles $z=k, k \in \mathbb{Z}$ and $z=-1/6$ and $z=-5/6$ is zero.  We thus have
$$\sum_{k=0}^{\infty} \left ( \frac1{6 k+1} - \frac1{6 k+5}\right )  =-\frac{\pi}{18} \left [\frac{\cot{\left ( -\frac{\pi}{6} \right )}}{\frac{4}{6}} + \frac{\cot{\left ( -\frac{5 \pi}{6} \right )}}{-\frac{4}{6}} \right ] = 2 \sqrt{3} \frac{\pi}{18} \cdot \frac{3}{2} = \frac{\pi}{2 \sqrt{3}}$$
