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For which values of $z$ does $$\sum_{n=1}^\infty \frac{\tan(nz)}{n^2}$$ converge? For which values of $z$ is the limiting function analytic?

One can show, as in this answer, that $$\left|\frac{e^{inz}-e^{-inz}}{e^{inz}+e^{-inz}}\right|$$ is bounded as $n\to \infty$, so long as $\text{Im}(z)\neq 0$. But the article above really does not discuss the case $\text{Im}(z)=0$, although it thinks it does. It doesn't deal with the poles at $\frac{(2k+1)\pi}{2}$, which can make some of the terms of the series undefined.

If $\text{Im}(z)=0$, obviously the estimate $$\left| \frac{e^{inz}-e^{-inz}}{e^{inz}+e^{-inz}} \right|\leq \frac{1+e^{2ny}}{|1-e^{2ny}|} $$

does not work. (Here $y=\text{Im}(z)$.) For $x\in \mathbb{R}$ of the form $j^2\frac{(2k+1)\pi}{2}$, there will be undefined terms.

Suppose there are no undefined terms. What can we say then about convergence? And in what way can we describe these singularities of the limiting function, corresponding to $x$ with undefined terms? Perhaps these points are not even isolated...

Eric Auld
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1 Answers1

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Following sos440's answer given here, we can show that this series diverges whenever the irrationality measure of $z/\pi$ is greater than $3$.

Let $\mu$ denote the irrationality measure of $\frac{z}{\pi}$. Then, for any $\eta<\mu$, there exists infinitely many $p,q$ such that $$\left|\frac{1}{\pi z}-\frac{2p+1}{2q}\right|\leq\frac{1}{q^{\eta}},$$ and so $$\left|qz-\frac{\pi}{2}-p\pi\right|\leq\frac{\pi}{q^{\eta-1}}.$$ Consequently, $$|\tan(qz)|=|\tan\left(\frac{\pi}{2}+\left(qz-\frac{\pi}{2}-p\pi\right)\right)\gg\frac{1}{\left|qz-\frac{\pi}{2}-p\pi\right|}\gg q^{\eta-1}.$$ Thus there exists a constant $C$ such that for infinitely many $q$ $$\frac{\tan(qz)}{q^{2}}\geq Cq^{\eta-3}.$$ If $\mu>3$ so that $\eta$ can be taken to equal $3$, then there are infinitely many terms in the series bounded away from $0$, which implies that it diverges.

This implies that if $z=\ell/\pi$ where $\ell$ is Liouville's number, then the series diverges.

For more details, see Wadim Zudilin's answer on MathOverflow concerning the convergence of $$\sum_{n=1}^\infty \frac{1}{n^3\sin^2(n)}.$$ His comment concerning quadratic irrationals can be used to prove that if $\frac{z}{\pi}$ is a quadratic irrational, then the series converges, and similarly if $z=\pi e$ then the series diverges.

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Eric Naslund
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  • But if $\mu=2$, the argument seems inconclusive, since you get infinitely many $q$ such that $|\tan(qz)/q^2|\geq C/q$, but then you can still have convergence (though not absolute convergence). Well, if I understand correctly :-) – Jean-Claude Arbaut Dec 23 '13 at 19:52
  • @Arbautic: The question of convergence when $\mu=2$ is very delicate. For quadratic irrationals, the denominator of the best convergents increases geometrically, and so the series converges. However this is not always the case. See Wadim Zudilin's answer here: http://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1 – Eric Naslund Dec 23 '13 at 20:01
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    Yes, I understand it's a difficult case, and indeed one cannot conclude only by this argument ;-) – Jean-Claude Arbaut Dec 23 '13 at 20:03
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    @EricNaslund Wow, this question led unexpectedly into quite deep water... :) – Eric Auld Dec 23 '13 at 20:25