Do Riemann's zeta-function's partial sums
∑Nn=1n−s converge conditionally for some value s=σ+it with σ≤1? (We must at least have t≠0 of course.)
Partial summation does not work because cos(tlogn) does not have bounded sums, but I wonder if perhaps at least for σ=1 and some t≠0 we may have convergence.
1st Edit: I insist that I am not interested in absolute convergence, which I understand. I really want to know if enough cancellation occurs in the complex powers n1+it, t≠0 for the ordered sequence of partial sums to converge—i.e. for the series to converge conditionally.
I guess that this issue may be related to elementary estimates used to prove the prime number theorem (like those of Erdős and Selberg)—even if none implies conditional convergence.
2nd Edit: To recap, conditional convergence at σ of a Dirichlet series ∑n≥1ann−s, with real an implies no pole on the real half-line at the right of σ so the abscissa of absolute and conditional convergence of the Dirichlet series representations (which is unique, a nontrivial result) for Riemann's ζ are the same, 1, i.e. the series does not converge conditionally for σ<1.
I will also mention that the Dirichlet series ∑n≥1(−1)nn−s has abscissa of conditional convergence 0 (therefore no pole at the right of 0), and dividing it by 21−s−1 we obtain ζ(s), so this is close to a Dirichlet series evaluation of ζ—which are known not to be practical computationally.
I could find interesting results in Tenenbaum's book on analytic number theory. I guess I will have to look at the heavy weight references, specialized on Riemann's zeta-function.
The case of σ=1 and t≠0 is still unsettled in the answers to this question, and in my mind.
3rd Edit: This question on mathoverflow seems to address exactly my question:
https://mathoverflow.net/questions/84097/divergence-of-dirichlet-series
The conclusion, there, is that the series diverges also for t≠0. This may be related to the existence of unbounded functions with bounded mean oscillation, like logt.
I'll read more about that and think about it.
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