[Math] Conditional convergence of Riemann’s $\zeta$’s series

calculusnumber theoryriemann-zeta

Do Riemann's zeta-function's partial sums
$\sum_{n=1}^N n^{-s}$ converge conditionally for some value $s=\sigma+it$ with $\sigma\le 1$? (We must at least have $t\ne 0$ of course.)

Partial summation does not work because $\cos(t\log n)$ does not have bounded sums, but I wonder if perhaps at least for $\sigma=1$ and some $t\ne 0$ 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 $n^{1+it}$, $t\ne 0$ 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 $\sigma$ of a Dirichlet series $\sum_{n\ge 1} a_nn^{-s}$, with real $a_n$ implies no pole on the real half-line at the right of $\sigma$ so the abscissa of absolute and conditional convergence of the Dirichlet series representations (which is unique, a nontrivial result) for Riemann's $\zeta$ are the same, $1$, i.e. the series does not converge conditionally for $\sigma<1$.

I will also mention that the Dirichlet series $\sum_{n\ge 1}(-1)^nn^{-s}$ has abscissa of conditional convergence $0$ (therefore no pole at the right of $0$), and dividing it by $2^{1-s}-1$ we obtain $\zeta(s)$, so this is close to a Dirichlet series evaluation of $\zeta$—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 $\sigma=1$ and $t\ne 0$ 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 $t\ne 0$. This may be related to the existence of unbounded functions with bounded mean oscillation, like $\log t$.

I'll read more about that and think about it.

Best Answer

$\zeta$ is a Dirichlet series. As power series have a radius of convergence, Dirichlet series have an abscissa of convergence --- they converge to the right of a vertical line, and diverge to the left of it. For $\zeta$, that abscissa is $\sigma=1$. There some discussion here.