Not quite a telescoping sum (2)

$\large \displaystyle S = \sum_{n=0}^{\infty} \left [ \frac{1}{12n+1} + \frac{1}{12n+5} - \frac{1}{12n+7} - \frac{1}{12n+11} \right ]$

$\large \displaystyle S = \sum_{n=0}^{\infty} \left [ \frac{1}{12n+1} - \frac{1}{12n+3} + \frac{1}{12n+5} - \frac{1}{12n+7} + \frac{1}{12n+9} - \frac{1}{12n+11} \right ] + \sum_{n=0}^{\infty} \left [ \frac{1}{12n+3} - \frac{1}{12n+9} \right ]$

$\large \displaystyle S = \sum_{n=0}^{\infty} \left [ \frac{(-1)^n}{2n+1} \right ] + \frac13 \sum_{n=0}^{\infty} \left [ \frac{1}{4n+1} - \frac{1}{4n+3} \right ]$

$\large \displaystyle S = \sum_{n=0}^{\infty} \left [ \frac{(-1)^n}{2n+1} \right ] + \frac13 \sum_{n=0}^{\infty} \left [ \frac{(-1)^n}{2n+1} \right ]$

$\large \displaystyle S = \frac43 \sum_{n=0}^{\infty} \left [ \frac{(-1)^n}{2n+1} \right ]$

By the Maclaurin Series of $\arctan$

$\large \displaystyle S = \frac43 \arctan(1)$

$\color{#20A900} \boxed {\large \displaystyle S = \frac{\pi}{3} }$

So:

$\color{#3D99F6} \boxed {\large \displaystyle a = 3}$

$\begin{aligned} S & = \sum_{n=0}^\infty \left(\frac 1{12n+1} + \frac 1{12n+5} - \frac 1{12n+7} - \frac 1{12n+11} \right) \\ & = 1 + \frac 15 - \frac 17 - \frac 1{11} + \frac 1{13} + \frac 1{17} - \frac 1{19} - \frac 1{23} + \cdots \\ & = 1 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \frac 1{11} + \frac 1{13} - \frac 1{15} + \cdots + \left( \frac 13 - \frac 19 + \frac 1{15} - \frac 1{21} + \frac 1{27} - \frac 1{33} + \cdots \right) \\ & = 1 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \frac 1{11} + \frac 1{13} - \frac 1{15} + \cdots + \frac 13 \left(1 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \frac 1{11} + \cdots \right) \\ & = \frac 43 \left(1 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \frac 1{11} + \frac 1{13} - \frac 1{15} + \cdots \right) \\ & =\frac 43 \tan^{-1} (1) = \frac 43 \times \frac \pi 4 = \frac \pi 3 \end{aligned}$

Therefore, $a=\boxed 3$ .

$\sum_{n=0}^{\infty}\left(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{12n+5}-\frac{1}{12n+7}\right) +\sum_{n=0}^{\infty}\left(\frac{1}{12n+1}-\frac{1}{12n+11}\right) \\=\frac{1}{12}\sum_{n=0}^{\infty}\left(\frac{1}{n+\frac{5}{12}}-\frac{1}{n+\frac{7}{12}}\right) +\frac{1}{12}\sum_{n=0}^{\infty}\left(\frac{1}{n+\frac{1}{12}}-\frac{1}{n+\frac{11}{12}}\right)=\frac{1}{12}\left(\psi^0\left(\frac{7}{12}\right)-\psi^0\left(\frac{5}{12}\right)+\psi^0\left(\frac{11}{12}\right)- \psi^0\left(\frac{1}{12}\right)\right)$ Using the relation $\psi(1-z)-\psi(z)= \pi\cot(\pi z)$ gives us $\frac{\pi}{12}\left(\cot\left(\frac{5\pi}{12}\right)+\cot\left(\frac{\pi}{12}\right)\right)=\frac{4\pi}{12}=\frac{\pi}{3}$ Since $\cot A +\cot B=\frac{\sin(A+B)}{\sin A\sin B}$ putting $A=\frac{\pi}{12}$ and $B=\frac{5\pi}{12}$ . We have $\cot\left(\frac{\pi}{12}\right)+\cot\left(\frac{5\pi}{12}\right)=\frac{\sin\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)}=\frac{1}{\sin\left(\frac{\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)}\\=\frac{2}{\sin\left(\frac{\pi}{6}\right)}=4$