# Conditional convergence of Riemann's $\zeta$'s series

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.

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 ErdÃ¶s and Selberg) -even if none implies conditional convergence.

$2$ND 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.

$3$RD EDIT: This question on mathoverflow seems to address exactly my question: http://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 $\log t$.

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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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As Gerry says, it diverges on the left of the abscissa of convergence, so it diverges for all $\sigma < 1$. – John Wordsworth Dec 4 '12 at 10:16
@draks, if you understood the names that Gerry wrote thn you shouldn't ask that, ergo: you seem to not understand some of them, so perhaps it'd be a good idea to google "Dirichlet series:, or go directly to en.wikipedia.org/wiki/Dirichlet_series – DonAntonio Dec 4 '12 at 10:26
ah sorry, I skipped the middle part of the answer. Sorry for the inconvenience. – draks ... Dec 4 '12 at 10:28
The page you point to does not address my question. Your answer only treats absolute convergenge, which I understand. The wiki page says: "absolute convergence and uniform convergence may occur in distinct half-planes" which I guess implies that Dirichlet series may converge conditionally on the left of their "abscissa of convergence" (which really refers to absolute convergence). My question remains: do the partial sums converge conditionally for some $\sigma=1$? – plm Dec 4 '12 at 11:37
My apologies --- things are a little more complicated than I thought --- see math.harvard.edu/archive/213b_spring_05/dirichlet_series.pdf – Gerry Myerson Dec 4 '12 at 12:03
$$s=1+ti\,\,,\,t\neq 0\Longrightarrow n^s=n\cdot n^{it}\Longrightarrow |n^|=n\Longrightarrow$$
$$=\sum_{n=1}^\infty\frac{1}{n^s}\,\,\,\text{diverges}$$
Your implications (assuming you meant $|n^s|$ where there is a typo) only show the sum does not converge absolutely, not that it does not converge conditionally. – plm Dec 4 '12 at 11:29
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 $1+it$, $t\ne 0$. – plm Dec 5 '12 at 11:14