$\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ and $ S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i}$ . Find $S_1 + S_2 $ I know to find sum of series using method of difference. I tried sum of write the term as (6n-1)(6n+1). i don't know how to proceed further.

Assuming that you know about the polygamma functions, first write
$$(36n^2-1)^2=(6n+1)^2(6n-1)^2$$ and perform a first partial fraction decomposition
$$\frac 2{(36n^2-1)^2}=\frac{1}{2 (6 n+1)}-\frac{1}{2 (6 n-1)}+\frac{1}{2 (6 n+1)^2}+\frac{1}{2 (6 n-1)^2}$$
Computing the partial sum
$$\Sigma_p=\sum_{n=1}^p\frac 2{(36n^2-1)^2}=$$ $$\Bigg[\frac{1}{12} \left(\psi ^{(0)}\left(p+\frac{7}{6}\right)-\psi
   ^{(0)}\left(\frac{7}{6}\right)\right) \Bigg]-\Bigg[\frac{1}{12} \left(\psi ^{(0)}\left(p+\frac{5}{6}\right)-\psi
   ^{(0)}\left(\frac{5}{6}\right)\right) \Bigg]+\Bigg[\frac{1}{72} \left(\psi ^{(1)}\left(\frac{7}{6}\right)-\psi
   ^{(1)}\left(p+\frac{7}{6}\right)\right) \Bigg]+\Bigg[\frac{1}{72} \left(\psi ^{(1)}\left(\frac{5}{6}\right)-\psi
   ^{(1)}\left(p+\frac{5}{6}\right)\right) \Bigg]$$
Now, using the asymptotics
$$\Sigma_p=\frac{1}{72} \left(6 \psi ^{(0)}\left(\frac{5}{6}\right)-6 \psi
   ^{(0)}\left(\frac{7}{6}\right)+\psi ^{(1)}\left(\frac{5}{6}\right)+\psi
   ^{(1)}\left(\frac{7}{6}\right)\right)-\frac{1}{1944 p^3}+O\left(\frac{1}{p^4}\right)$$ Thus
$$S_2=\frac{1}{72} \left(6 \psi ^{(0)}\left(\frac{5}{6}\right)-6 \psi
   ^{(0)}\left(\frac{7}{6}\right)+\psi ^{(1)}\left(\frac{5}{6}\right)+\psi
   ^{(1)}\left(\frac{7}{6}\right)\right)=-1+\frac{\pi }{4 \sqrt{3}}+\frac{\pi ^2}{18}$$
$$S_1+S_2=\frac{2 \pi ^2}{9}+\frac{\pi }{4 \sqrt{3}}-1\sim 1.64670 $$
Obviously, this is not the same result as in other answers.
I must say that I do not understand how, adding positive terms to $S_1$, we could have a vlaue smaller then $\frac{ \pi ^2}{6} \sim 1.64493$.
