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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.

3 Answers 3

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$s=1+ti\,\,,\,t\neq 0\Longrightarrow n^s=n\cdot n^{it}\Longrightarrow |n^|=n\Longrightarrow$

$=\sum_{n=1}^\infty\frac{1}{n^s}\,\,\,\text{diverges}$

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    Well it does not explain things very clearly but I could find that $\zeta$ does not converge conditionally at the left of the abscissa of absolute convergence. It does not address points $1+it$, $t\ne 0$.2012-12-05
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$\zeta$ 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.

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    The first result in that link, on uniform convergence, is really tight, it implies that there is no pole on the real half-line at the right of the abscissa of conditional convergence, for any Dirichlet series. So $\zeta$ does not converge conditionally for \sigma<1. That does not settle $\sigma=1$. (I am going to read other answers.)2019-04-07
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I was looking for an answer to this question and was surprised to find that the domain of conditional convergence was the same as the domain of absolute convergence, i.e. $Re(z) > 1$ (see https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ for example).

Intuitively this feels wrong because if $z = a + ib$ with $0 < a < 1$ and $b \neq 0$, then $n^a$ decreases towards $0$ and $n^{ib}$ rotates by $b\ log(n)$. So we'd expect that the partial sums would spiral towards a limit because of the cancellations introduced by the rotation.

I wrote a small program to test it and it seems to confirm that the series converges conditionally in the $0 < Re(z) <= 1$ band. I tested on the critical line to see if conditional convergence would detect the zeros and it did.

https://gist.github.com/bjouhier/77fd50e1d43be2e813abd433e65d31f6

This is far from a from a formal proof but it feels like the series converges conditionally in the $0 < Re(z) <= 1, Im(z) \neq 0$ domain

Post-mortem: intuition was wrong: the angle of rotation between successive terms is $b(log(n+1)-log(n))$, i.e. $b/n$. So rotation slows down as $n$ increases, cancellation take more and more terms, and the spiral does not converge. Domain of conditional convergence is $Re(z) > 1$, same as absolute convergence. See comments.

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    Got it. My intuition was wrong because rotation slows down as $n$ grows (it is $(log(n+1) - log(n))b$ between successive terms). So cancellation will happen less and less often and partial sums will diverge. My program only exercises the space were $b$ dominates $(log(n+1) -log(n))$.2016-05-06