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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: 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 $\log t$.

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

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2 Answers 2

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

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

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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
    
True as I thought that's what you meant but, obviously, I misread. Anyway, the abscissa of convergence of Dirichlet Series covers this case, too. –  DonAntonio Dec 4 '12 at 11:31
    
I do not think that the abscissa of convergence refers to issues of conditional convergence neither that it implies much (straightforwardly) about my question. As mentioned above it seems that uniform convergence of the partial sums may occur at the left of the abscissa of (absolute) convergence. Thanks for the attempt though. –  plm Dec 4 '12 at 11:53
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I think that after reading the following things will clear up: mathreference.com/lc-z,cond.html –  DonAntonio Dec 4 '12 at 12:19
    
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

$\zeta$ 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 $\sigma < 1$. –  Old John 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
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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

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