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We know that $\sum_{n=1}^\infty \frac{z^n}{n}$ converges if $|z|=1$ except when $z=1$. From this it is simple to construct another series that converges whenever $z$ is a $m$th root of unity, where $m < M$ for some $M$, with $\sum_{m=1}^M \sum_{n=1}^\infty \frac{z^{mn}}{n}$.

Is it possible to construct a sum that diverges whenever $z$ is any root of unity? I suspect that $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{z^{mn}}{n}$ is not actually a convergent series and thus naively taking the limit of the above series will not work.

Update: How about the series $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{z^{mn}}{nm}$? This one diverges for $\arg z/2\pi$ irrational, but I expect that it should converge otherwise. If not, how about $m!$ in the denominator?

Integrand
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SKK
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  • If a power series converges on all irrationals, then it should converge to all rationals inside the circle of convergence. – Anupam Nov 23 '18 at 08:51
  • Sorry, poor explanation; a power series that converges on all |z|=1 with (arg z )/2pi irrational. – SKK Nov 23 '18 at 15:41

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Let $z= \exp(2\pi i x)$ for $x\in [0,1]$ and consider $${ \sum_{n=1}^{\infty} \frac{ z^{n!}}{n}=\sum_{n=1}^{\infty} \frac{\exp(2\pi i n! x)}{n} }.$$ If $z$ is an $m^{th}$ root of unity, i.e. if $x$ is rational, there is an $n$ such that $m|n!$, in which case the sum is eventually comparable to the divergent harmonic series. However, as explained here, we can use the high-powered Carleson's theorem to show the series converges pointwise a.e. on $[0,1]$.

Integrand
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