Let, the sum be $S$, then we can make the following identity because $\text{sgn_1}$ of $2,3$ is defined to be $1$ modulo $4$. 

 The identity : $$S=1+\frac{S}{2}+\frac{S}{3}-\frac{S}{6}+\left(\sum_{n\geq 0}\frac{1}{12n+1}-\sum_{n\geq 1}\frac{1}{12n-1}\right)+\left(\sum_{n\geq 0}\frac{1}{12n+5}-\sum_{n\geq 1}\frac{1}{12n-5}\right)$$

This gives $\frac{S}{3}=(1+\frac{1}{5})+\frac{1}{12^2}\left(\sum_{n=1}^{\infty} \frac{2}{(\frac{1}{12})^2-n^2}+\sum_{n=1}^{\infty} \frac{
10}{(\frac{5}{12})^2-n^2}\right)$

or, $\frac{S}{3}=\frac{\pi}{12}\left(\text{cot}(\frac{\pi}{12})+\text{cot}(\frac{5\pi}{12})\right)$

This gives $S=\pi$