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.
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.
 A: $\zeta$ 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. 
A: 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.
A: $$s=1+ti\,\,,\,t\neq 0\Longrightarrow n^s=n\cdot n^{it}\Longrightarrow |n^|=n\Longrightarrow$$
$$=\sum_{n=1}^\infty\frac{1}{n^s}\,\,\,\text{diverges}$$
