TITLE: Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$
QUESTION [17 upvotes]: 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?

REPLY [6 votes]: My first thought did not use the identity in the question, but used identity $(7)$ proven in this answer,
$$
\sum_{k\in\mathbb{Z}}\frac1{z+k}=\pi\cot(\pi z)\tag{1}
$$
so that
$$
\begin{align}
\sum_{k\in\mathbb{Z}}\frac1{1+6k}
&=\frac16\sum_{k\in\mathbb{Z}}\frac1{\frac16+k}\\
&=\frac16\pi\cot\left(\frac\pi6\right)\\[6pt]
&=\frac\pi6\sqrt3\\[6pt]
&=\frac\pi{2\sqrt3}\tag{2}
\end{align}
$$

To use the identity in the question,
$$
\frac\pi2-\frac x2=\sum_1^\infty\frac{\sin(nx)}n\tag{3}
$$
it appears that we could consider $x=\frac\pi3$:
$$
\begin{align}
\frac\pi2-\frac\pi6
&=\sum_{n=1}^\infty\frac{\sin\left(\frac{n\pi}3\right)}{n}\\
&=\frac{\sqrt3}2\left(\color{#C00000}{\frac11}\color{#0000F0}{+\frac12-\frac14}\color{#C00000}{-\frac15+\frac17}\color{#0000F0}{+\frac18-\frac1{10}}\color{#C00000}{-\frac1{11}+\frac1{13}}+\dots\right)\\
&=\frac{\sqrt3}2(\color{#C00000}{A}+\color{#0000F0}{B})\tag{4}
\end{align}
$$
Note that the sum we seek is $A$. It is an easy exercise to verify that the series for both $A$ and $B$ converge (using The Alternating Series Test).
Now we have that $(A+B)+2B=2A$:
$$
\begin{array}{rl}
\displaystyle A+B=&\displaystyle\frac11+\frac12-\frac14-\frac15+\frac17+\frac18-\frac1{10}-\frac1{11}+\frac1{13}+\dots\\
\displaystyle 2B=&\displaystyle\frac11-\frac12+\frac14-\frac15+\frac17-\frac18+\frac1{10}-\frac1{11}+\frac1{13}+\dots\\
\displaystyle 2A=&\displaystyle\frac21\phantom{+\frac02-\frac04}\ \ -\frac25+\frac27\phantom{+\frac08-\frac0{10}}\ \ -\frac2{11}+\frac2{13}+\dots\tag{5}
\end{array}
$$
Thus, $A=3B$. Using this in $(4)$ gives
$$
\frac\pi3=\frac{\sqrt3}2\left(A+\frac13A\right)\tag{6}
$$
and solving for $A$ yields
$$
A=\frac\pi{2\sqrt3}\tag{7}
$$