# Sum of reciprocals of squares of the form $3n+1$?

What is $\sum\limits_{n=0}^\infty \frac1{(3n+1)^2}$?

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The sum is easily expressed in terms of the Lerch transcendent: $\dfrac19 \Phi\left(1,2,\dfrac13\right)$. There is an alternative expression in terms of the trigamma function: $\dfrac19\psi^{(1)}\left(\dfrac13\right)$, but deriving that form is a bit more complicated to do... –  Guess who it is. Feb 8 '12 at 17:18
The title and the question are asking two different things: the question asks for squares of (numbers that are of the form $3n+1$), whereas I at least parsed the title as asking for (squares of numbers) that are of the form $3n+1$; the sum in (my interpretation of) the title has an explicit elementary form, as opposed to the sum of the question... –  Steven Stadnicki Feb 8 '12 at 17:54
See mathoverflow.net/questions/87348/… by, one assumes, the same person. –  Will Jagy Feb 8 '12 at 21:03

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,

$$\sum_{k=0}^\infty \frac1{(3k+1)^2}=\frac19\Phi\left(1,2,\frac13\right)=\frac19\psi^{(1)}\left(\frac13\right)$$

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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 .

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Some numbers are negative, so the title (not the actual statement, though) may refer to this one, easier to do: $$\sum_{k=-\infty}^\infty \frac{1}{(3k+1)^2} = \frac{4\pi^2}{27}$$

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The sum is $$\frac19{\Psi(1,\frac13)}$$

where $\Psi$ is the Poligamma function

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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.

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