Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

$\zeta$ 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.

share|cite|improve this answer
    
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
    
"They converge to the right of a vertical line, and diverge to the left of it." If they converge conditionally, they converge --- but I wrote they diverge, and that means they diverge, and don't converge, period. "uniform convergence" has nothing to do with conditional convergence, it's another topic altogether. I grant you that this doesn't account for $\sigma=1$, but it does account for $\sigma\lt1$. – Gerry Myerson Dec 4 '12 at 11:56
2  
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
    
Great paper (by You Tong Siu I remark), I do not have time to read it much now, but it does seem that $\zeta$ converges on and at the left of its "abscissa of convergence". – plm Dec 4 '12 at 12:20
    
Well, $(1-(1/2^{s-1}))\zeta(s)$ has that property, but as $\zeta$ itself diverges at $s=1$, that would seem to settle the matter of convergence for $\sigma\le1$. – Gerry Myerson Dec 4 '12 at 22:54

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

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

share|cite|improve this answer
    
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
2  
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

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.

share|cite|improve this answer
    
No, if the series converges (conditionally) for some $z_0$, then it converges (conditionally) for all $z$ with $\operatorname{Re} z > \operatorname{Re} z_0$. (Even that the sequence of partial sums is bounded would suffice.) What might be the case is that the series is conditionally convergent for some $1 + it$. – Daniel Fischer May 6 at 20:21
    
The series does not converge on the real line when $x <= 1$ but it looks like it converges (conditionally) to the left of the $Re(z) = 1$ line when the imaginary part is not $0$. Where is the proof that convergence on $z0$ implies convergence on all $z$ such that $Re(z) > Re(z0)$? – Bruno Jouhier May 6 at 20:39
    
It may look so, but it isn't so. The divergence is slow, however, so you need a) many terms and b) high precision to detect it. I linked to a proof in my first comment, the word "No" is a link. – Daniel Fischer May 6 at 20:42
    
Proof is backwards. If series converges for $s$ real then it converges for all $z$ with $Re(z) = s$. This does not say anything about the case where it converges for $z$ outside the real line. – Bruno Jouhier May 6 at 21:03
    
We can absorb the $n^{-it}$ into the coefficients. If $$P_N := \sum_{n = 1}^{N} \frac{a_n}{n^{z_0}}$$ is a bounded sequence, consider the series with coefficients $b_n = a_n\cdot n^{-z_0}$. – Daniel Fischer May 6 at 21:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.