Sum of two infinite series: Σ[1/(36r^2-1)+2/(36r^2-1)^2]

[juantheron]

Evaluation of [math]\displaystyle \sum_{r=1}^\infty \left(\frac{1}{36r^2-1}+\frac{2}{(36r^2-1)^2}\right)[/math]

 

[Olinguito]

Spoiler

In partial fractions,
$$\frac{1}{36r^2-1}+\frac{2}{(36r^2-1)^2}\ =\ \frac12\left[\frac1{(6r-1)^2}+\frac1{(6r+1)^2}\right].$$
So the sum is equal to
$$\frac12\left(\frac1{5^2}+\frac1{7^2}+\frac1{11^2}+\frac1{13^2}+\cdots\right)$$

$=\ \frac12(A-B-C+D)-\frac12$

where

$\displaystyle A\ =\ \frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\cdots\ =\ \frac{\pi^2}6$,

$\displaystyle B\ =\ \frac1{2^2}+\frac1{4^2}+\frac1{6^2}+\cdots\ =\ \frac14\cdot\frac{\pi^2}6$,

$\displaystyle C\ =\ \frac1{3^2}+\frac1{6^2}+\frac1{9^2}+\cdots\ =\ \frac19\cdot\frac{\pi^2}6$,

$\displaystyle D\ =\ \frac1{6^2}+\frac1{12^2}+\frac1{18^2}+\cdots\ =\ \frac1{36}\cdot\frac{\pi^2}6$.

Hence:
$$\sum_{r=1}^\infty\left[\frac1{36r^2-1}+\frac2{(36r^2-1)^2}\right]\ =\ \frac12\left(1-\frac14-\frac19+\frac1{36}\right)\frac{\pi^2}6-\frac12\ =\ \boxed{\frac{\pi^2}{18}-\frac12}.$$

 

[juantheron]

Thanks https://mathhelpboards.com/members/olinguito/. My solution is almost same as yours.