<s>[INST] How to take every third element in a series? $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$  or  $\frac{\pi^2}{6}$
What if we take every 3rd term and add them up? 
A = $ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ \frac{1}{2^2} + \frac{1}{5^2} + \frac{1}{8^2} + \cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ \frac{1}{1^2} + \frac{1}{4^2} + \frac{1}{7^2} + \cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
 [/INST] Note that we have
$$\psi'(z)=\sum_{n=0}^\infty \frac{1}{(n+z)^2}$$
where $\psi'(z)$ is the derivative of the digamma function.  Hence, we can write
$$\sum_{n=0}^\infty \frac{1}{(3n+1)^2}=\frac19 \psi'(1/3)$$
and 
$$\sum_{n=0}^\infty \frac{1}{(3n+2)^2}=\frac19 \psi'(2/3)$$
Interestingly, since we have
$$\sum_{n=0}^\infty \left(\frac1{(3n+3)^2}+\frac1{(3n+2)^2}+\frac1{(3n+1)^2}\right)=\frac{\pi^2}{6}$$
we find that
$$\psi'(1/3)+\psi'(2/3) = 4\pi^2/3$$
</s>