Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$

Solution 1:

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 $$ A=\frac\pi{2\sqrt3}\tag{7} $$

Solution 2:

In the figure below, in which the trigonometric circle has been divided into six equal parts, we can see clearly that for $x= \frac {\pi}{3}$ the following equalities are verified:

$\sin( \frac{n\pi}{3})=\frac{\sqrt3}{2}$ for $n=1,2,7,8,……, 1+6n,2+6n,…..$

$\sin( \frac{n\pi}{3})=\frac{-\sqrt3}{2}$ for $n=4,5,10,11,......,4+6n,5+6n$

$\sin\left( \frac{n\pi}{3}\right)=0$ for $n=3,6,9,12,……,3n$

Hence, for $x= \frac {\pi}{3}$
$$\frac {\pi}{2} - \frac {\pi}{6}=\frac{\pi}{3}$$

$$\frac{\pi}{3}=\frac{\sqrt 3}{2}\sum_{n=1}^\infty \left[\frac{1}{1+6n} +\frac{1}{2+6n}-\frac{1}{4+6n}-\frac{1}{5+6n}\right]$$ We have $$\frac{\pi}{3}=\frac{\sqrt 3}{2}A\iff \frac{\pi}{2\sqrt3}=\frac{3}{4}A=(1-\frac14)A$$ From which we see the odd-denominator terms $$\frac{1}{1+6n} -\frac{1}{5+6n}$$ remain unchanged and the even-denominator terms $$\frac{1}{2+6n} -\frac{1}{4+6n}-\frac{1}{4+24n}-\frac{1}{8+24n}+\frac{1}{16+24n}+\frac{1}{20+24n}$$

give a telescopic series of total sum equal to zero.

(Note that $4+24n=4+6(4n)$;$8+24n=2+6(4n+1)$; $16+24n=4+6(4n+2)$;$20+24n=2+6(4n+3)$; we obtain $$0=\frac12-\left(\frac14+\frac14\right) = \frac18-\frac18=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{20}\right)=\frac{1}{14}-\left(\frac{1}{28}+\frac{1}{28}\right)$$ an so on.

Thus we end with the addition of the terms $$\frac{1}{1+6n} -\frac{1}{5+6n}$$