4
$\begingroup$

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

  • 1
    See http://mathoverflow.net/questions/87348/sum-of-reciprocals-of-squares-of-integers-congruent-to-1-mod-3 by, one assumes, the same person.2012-02-08

5 Answers 5

8

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,

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

2

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 .

2

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} $

1

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

where $\Psi$ is the Poligamma function

1

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.