Knowing that $$\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$$
and $$S_i=\sum_{k=1}^{\infty}\frac{i}{(36k^2-1)^i}$$
Find value of $S_1+S_2$
i tried splitting:
$$\frac{1}{36k^2-1}=\frac{1}{2}\times \left(\frac{1}{6k-1}-\frac{1}{6k+1}\right)$$
But no idea there after
First of all note the following
$$S_1+S_2=\sum_{k=1}^\infty\frac1{36k^2-1}+\sum_{k=1}^\infty\frac2{(36k^2-1)^2}=\frac12\sum_{k=1}^\infty \left[\frac1{(6k-1)^2}+\frac1{(6k+1)^2}\right]=S$$
which immediately follows from partial fraction decomposition and rearranging. Note that the latter is legitimate due the absolute convergence of the series. In order to evaluate the $S$ we rewrite it as
$$S=\frac12\sum_{k=1}^\infty \frac1{(6k-1)^2}+\frac1{(6k+1)^2}=-\frac12+\frac12\sum_{k=0}^\infty\left[\frac1{(6k+1)^2}+\frac1{(6k+5)^2}\right]$$
Now it is more clear what is going on here! Precisely we want to eliminate all numbers of the form $6k,6k+2,6k+3,6k+4$. Thus, lets us consider the reciprocal sum of all integers arranged in the aforementioned form
$$\sum_{k=1}^\infty\frac1{k^2}=\sum_{k=1}^\infty\frac1{(6k)^2}+\sum_{k=0}^\infty\frac1{(6k+1)^2}+\frac1{(6k+2)^2}+\frac1{(6k+3)^2}+\frac1{(6k+4)^2}+\frac1{(6k+5)^2}$$
The rest is algebraic manipulation. Thus, we get the following
$$\small\begin{align*} \sum_{k=1}^\infty\frac1{k^2}&=\sum_{k=1}^\infty\frac1{(6k)^2}+\sum_{k=0}^\infty\frac1{(6k+1)^2}+\frac1{(6k+2)^2}+\frac1{(6k+3)^2}+\frac1{(6k+4)^2}+\frac1{(6k+5)^2}\\ \sum_{k=1}^\infty\frac1{k^2}&=\sum_{k=1}^\infty\frac1{(6k)^2}+\sum_{k=0}^\infty\left[\frac1{(6k+1)^2}+\frac1{(6k+5)^2}\right]+\frac1{(6k+2)^2}+\frac1{(6k+3)^2}+\frac1{(6k+4)^2}\\ \sum_{k=1}^\infty\frac1{k^2}&=\frac1{36}\sum_{k=1}^\infty\frac1{k^2}+[2S+1]+\sum_{k=1}^\infty\frac19\left[\frac1{(2k+1)^2}\right]+\frac14\left[\frac1{(3k+1)^2}+\frac1{(3k+2)^2}\right]\\ \frac{35}{36}\sum_{k=1}^\infty\frac1{k^2}&=[2S+1]+\frac19\sum_{k=0}^\infty\left[\frac1{(2k+1)^2}\right]+\frac14\sum_{k=0}^\infty\left[\frac1{(3k+1)^2}+\frac1{(3k+2)^2}\right]\\ \frac{35}{36}\sum_{k=1}^\infty\frac1{k^2}&=[2S+1]+\frac19\left[\sum_{k=1}^\infty\frac1{k^2}-\sum_{k=1}^\infty\frac1{(2k)^2}\right]+\frac14\left[\sum_{k=1}^\infty\frac1{k^2}-\sum_{k=1}^\infty\frac1{(3k)^2}\right]\\ \frac{35}{36}\sum_{k=1}^\infty\frac1{k^2}&=[2S+1]+\frac19\left[\frac34\sum_{k=1}^\infty\frac1{k^2}\right]+\frac14\left[\frac89\sum_{k=1}^\infty\frac1{k^2}\right]\\ \frac23\sum_{k=1}^\infty\frac1{k^2}&=[2S+1] \end{align*}$$
$$\therefore~S=S_1+S_2=-\frac12+\frac12\sum_{k=0}^\infty \frac1{(6k+1)^2}+\frac1{(6k+5)^2}=\frac{\pi^2}{18}-\frac12$$
