# Convergence of $\sum_{n=1}^\infty \frac{\psi^{(1)}(1-n/\pi)}{n^3}$

Related to an open problem about another series.

Set

$$A= \sum_{n=1}^\infty \frac{\psi^{(1)}(1-n/\pi)}{n^3}$$

where $\psi^{(n)}(k)$ is the polygamma function.

Does $A$ converge?

The related series $\sum_{n=1}^\infty \frac{\psi^{(1)}(n/\pi)}{n^3}$ converges.

According to Maple 13: $\lim_{n \to \infty} \frac{\frac{\psi^{(1)}(1-n/\pi)}{n^3}}{\frac{1}{n^2}}=0$. If this is true it appears to give convergence. The series expansion at infinity involves $\cot$ so I suppose Maple's result might be wrong.

What other CASes say about the series?

I suppose $A$ converges iff $\sum_{n=1}^\infty \cot^2(n)/n^3$ converges.

Experimentally $A$ is about $278.73$.

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I think you are right. It is a question of how close multiples of $\pi$ can be to an integer. I think $$\lim_{n \to \infty} \frac{\frac{\psi^{(1)}(1-n/\pi)}{n^3}}{\frac{1}{n^2}}=0$$ is false if $n$ runs over the reals (which is what Maple assumes unless you somehow tell it otherwise). I could not get Maple 17 to tell me that the limit is zero. – Gerald Edgar Sep 16 '13 at 14:03
If we expand the polygamma function, your sum can be written $$A=\pi^2\sum_{q\ge1\atop p\ge1} q^{-2} p^{-3}\Big(\frac{p}{q}-\pi\Big)^{-2}.$$ A first quick check is to try an estimate in terms of the measure of irrationality of $\pi$ (at a first glance, it is not sufficient). – Pietro Majer Sep 16 '13 at 14:40
Or also $$A=\sum_{p\ge1\atop q\ge1} q^{-5}\Big(\frac{p}{q}-\frac{1}{\pi}\Big)^{-2}.$$ – Pietro Majer Sep 16 '13 at 15:06
Stopple, this is OK for OP's "related series" where the argument of the $\Psi^{1}$ goes to $+\infty$ but in $A$ it goes to $-\infty$, and all negative integers are poles of order 2 of $\Psi^{1}$. Check Gerald's comment. – Pietro Majer Sep 16 '13 at 15:28
@PietroMajer since this is an problem, partial answer is interesting to me. I am surprised I got $\cot$ in the argument... If convergence implies something other than $\mu(\pi) \le 2.5$ I will be surprised. – joro Sep 16 '13 at 15:50