Sum of reciprocals of squares of the form $3n+1$? What is $\sum\limits_{n=0}^\infty \frac1{(3n+1)^2}$?
 A: To expand on my comment, two functions here are relevant: the Lerch transcendent
$$\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(k+a)^s}$$
and the polygamma function
$$\psi^{(k)}(z)=\frac{\mathrm d^{k+1}}{\mathrm dz^{k+1}}\log\Gamma(z)=(-1)^{k+1}k!\sum_{j=0}^\infty \frac1{(z+j)^{k+1}}$$
where the series expression can be easily derived from differentiating the gamma function relation $\Gamma(z+1)=z\Gamma(z)$ an appropriate number of times
$$\psi^{(k)}(z+1)=\psi^{(k)}(z)+\frac{(-1)^k k!}{z^{k+1}}$$
and recursing as needed.
Comparing these definitions with the series at hand, we find that
$$\Phi\left(1,2,\frac13\right)=\sum_{k=0}^\infty \frac1{(k+1/3)^2}$$
which almost resembles the OP's series, save for a multiplicative factor:
$$\frac19\Phi\left(1,2,\frac13\right)=\sum_{k=0}^\infty \frac1{9(k+1/3)^2}=\sum_{k=0}^\infty \frac1{(3k+1)^2}$$
For the polygamma route, we specialize here to the trigamma case:
$$\psi^{(1)}(z)=\sum_{j=0}^\infty \frac1{(z+j)^2}$$
Letting $z=\frac13$, we have
$$\psi^{(1)}\left(\frac13\right)=\sum_{j=0}^\infty \frac1{(j+1/3)^2}$$
and we again see something familiar. Thus,
$$\sum_{k=0}^\infty \frac1{(3k+1)^2}=\frac19\Phi\left(1,2,\frac13\right)=\frac19\psi^{(1)}\left(\frac13\right)$$
A: Some numbers are negative, so the title (not the actual statement, though) may refer to this one, easier to do:
$$
\sum_{k=-\infty}^\infty \frac{1}{(3k+1)^2} = \frac{4\pi^2}{27}
$$
A: According to Maple solution is given by :
$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)^2} = \frac{1}{9} \Psi\left(1,\frac{1}{3}\right)$$
where $\Psi\left(1,\frac{1}{3}\right)$ is  polygamma function .
A: The sum is $$\frac19{\Psi(1,\frac13)}$$
where $\Psi$ is the Poligamma function
A: The sum can express 
$$\begin{align}
  & \sum\limits_{n=0}^{\infty }{\frac{1}{{{(3n+1)}^{2}}}} \\ 
 & =\frac{1}{9}\left[ \frac{{{\Gamma }''}(1/3)}{\Gamma (1/3)}-{{\left( \frac{{\Gamma }'(1/3)}{\Gamma (1/3)} \right)}^{2}} \right] \\ 
 & =1+\frac{1}{9}\int_{0}^{\infty }{\frac{t{{\mathrm{e}}^{-t/3}}}{{{\mathrm{e}}^{t}}-1}\mathrm{d}t} \\ 
\end{align}$$
But it can't find the specific value.
