# Evaluating $1/2^2 + 1/5^2 + 1/8^2 + 1/11^2 + \ldots$

Had been having fun evaluating positive integer powers of the reciprocals of the triangular and hexagonal numbers.

Currently, I'm trying to do the same for the pentagonal numbers, but I've run into a wall trying to evaluate the above series.

(BTW, if the "skip" had been every fourth number instead of every third, I can get expressions in terms of $\zeta(2)$ and Catalan's constant.)

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Mathematica spits out a formula involving the polygamma function, so I don't think you'll get a nicer expression than that. – TMM Mar 6 '12 at 19:21

As noted, we have only this in terms of the trigamma function $$\sum_{n = 0}^{\infty} \frac{1}{(3 n + 2)^{2}} = \frac{1}{9}\Psi' \Bigl(\frac{2}{3}\Bigr)$$ But there is this nicer one: $$\sum_{n = -\infty}^{\infty} \frac{1}{(3 n + 2)^{2}} = \frac{4\pi^2}{27}$$
Nice! So now all that's left is to find the value of $\sum_{n=-\infty}^{-1} \frac{1}{(3n+2)^2}$ and subtract it from your result ;) – TMM Mar 6 '12 at 19:58
@TMM: I can do that! It's $\frac{4\pi^2}{27} -\frac{1}{9}\Psi \Bigl(\frac{2}{3}\Bigr)$. – Stephen Mar 6 '12 at 20:07
Actually I think that should be $\frac{1}{9} \Psi'\left(\frac{2}{3}\right)$. – Robert Israel Mar 6 '12 at 20:21
Alternatively, the sum is expressible in terms of Hurwitz zeta: $$\frac19\zeta\left(2,\frac23\right)$$ – J. M. Jul 17 '12 at 4:56