# On the domain of convergence of usual definition of Riemann zeta function [closed]

If we define Riemann zeta function as it is usually defined so that $$\zeta (s)=\sum_{n=1}^{\infty}n^{-s}=\sum_{n=1}^{\infty}e^{-s\ln n}$$ then we can rewrite it as:

$$\zeta (s)=\sum_{n=1}^{\infty}\frac {1}{n^x}\cos(y\ln n) - i\sum_{n=1}^{\infty}\frac {1}{n^x} \sin(y\ln n)=w_1(x,y)-iw_2(x,y)$$:

It is known that for $y=0$ both $w_1$ and $w_2$ converge if $x \in (1,+\infty)$ because when $y=0$ we get domain of absolute convergence of $\zeta$.

Now, the question is what about the set of all $y$ such that both $w_1$ and $w_2$ converge for all $x$ in some interval (where the interval is dependent upon $y$ and is pushed beyond $\Re (s) >1$), in other words, the set of all $y$ such that $\zeta (s)$ defined as $\zeta (s)=\sum_{n=1}^{\infty} e^{-s\ln n}$ converges for all $x$ in some interval, so i raise the following question:

Is it true that for every $y\ne 0$ there is $0<\varepsilon<1$ (or we could write $\varepsilon(y)$ because of the dependence of $\varepsilon$ on $y$) such that $w_1$ and $w_2$ both converge for all $x \in (1-\varepsilon (y),+\infty)$?

So the question is about pushing the boundary of absolute convergence to the natural boundary of convergence of the formula that is used almost everywhere as a definition of Riemann zeta for $\Re (s) >1$.

• No, it's not true. When $y \neq 0$, the real and imaginary parts are basically power series in $e^{-x}$ (with non-integral powers) whose coefficients tend to be near 1. Note that $w_2$ is identically zero on the real line. – S. Carnahan Sep 23 '14 at 9:38
• But when $y \neq 0$ then real and imaginary parts have nonvanishing sine and cosine terms which for every $y \neq 0$ and different $n$ may in some way oscillate in $+$ and $-$ signs so they could make the series convergent, right? – Ante Sep 23 '14 at 9:42
• I see, you are asking for a non-uniform pointwise convergence of a rapidly oscillating sum, for all points in an open interval. I'm afraid I can't help you there. – S. Carnahan Sep 23 '14 at 10:03
• This is a Dirichlet series, and it diverges for $s=1$, hence it diverges for all $s$ with $\operatorname{Re}s<1$. – Emil Jeřábek Sep 23 '14 at 10:05
• @ Emil Jeřábek It diverges for $s=1$ because $1$ is a complex number whose imaginary part is zero and because harmonic series is divergent series, I do not see the reason why $w_1$ and $w_2$ should diverge for every complex number $s$ such that $\operatorname{Re}s<1$ if imaginary part is not zero? – Ante Sep 23 '14 at 10:15

I would have put this in a comment, but I don't have enough reputation to comment. So here it is: If a Dirichlet-Reihe $L(s)$ converges for some $s=s_0$, no matter if absolute or not, then it converges locally uniformly in every domain of the form $\{ s_0+z: |\arg(z)|<\frac\pi 2-\delta\}$, $\delta>0$ and therefore represents a holomorphic function in the domain $\Re(s)>\Re(s_0)$. As the Riemann zeta function has a pole at $s=1$, the series never converges for $\Re(s)<1$.
• But how do you know that there is no pair $(x,y)$ such that $y\ne 0$ and $x \in (1-\varepsilon, + \infty)$ and $0< \varepsilon <1$ such that both $w_1$ and $w_2$ converge, no matter absolutely or not? – Ante Sep 23 '14 at 10:31
• because then the Dirichlet-series would converge in the point $s_0=x+iy$. – Doug Sep 23 '14 at 10:50
• One more comment: the Dirichlet series argument does not determine for which points $1+it$ at the line of convergence the series converges. For Riemann zeta and similar series, mathoverflow.net/questions/84097 shows the answer is never. That is, the $\zeta(s)$ series only converges for $\operatorname{Re}s>1$. – Emil Jeřábek Sep 23 '14 at 11:51