# Riemann Zeta Function On Line Re(s)=1

I am having trouble thinking about this. Since the Riemann Zeta Function is analytic everywhere except at $s=1$, it follows that it is continuous on the real line $Re(s)=1$ except at $s=1$. Now, the Riemann Zeta Function is defined by a converging series for $Re(s)>1$ and this series does not converge for any other values of $s$. Hence, wouldn't the limit as $Re(s) \rightarrow 1$ be infinity? (due to divergence) But that cannot be, since the Riemann Zeta Function is supposed to be defined everywhere except for a pole at $s=1$. What am I missing?

• It is not clear what you are asking. Here $\frac{1}{z}$ has a pole at the origin and the limit in question is $\infty$. Where's the problem? – Mikhail Katz Aug 4 '14 at 10:35
• Ok. Stated differently: the value of, say, R(1+i), where with R I denote the RZF, exists, since it is defined everywhere except at s=1. Also, this value must equal the limit of R(x+1) as x tends to 1, since the function is continuous there ( has to be, it is analytic). But that limit, approached from the right side, diverges, since the infinite sum which defines the RZF for Re(s)>1 diverges everywhere else. – Asier Calbet Aug 4 '14 at 10:39
• Did you mean "from the left side" in your comment? – Mikhail Katz Aug 4 '14 at 10:41
• No, from the right, since on the right the function is defined by the converging and infinite series. – Asier Calbet Aug 4 '14 at 10:41
• The series may not be uniformly convergent for $Re(s)>1$ so you can't necessarily compare the value of the function and the value of the series when $Re(s)=1$. – Mikhail Katz Aug 4 '14 at 10:56

By way of comparison, think about the geometric series $$f(z)=\sum_{k=0}^\infty z^k$$ for $|z|<1$. This series has an analytic continuation as $$f(z)=\frac{1}{1-z}$$ for all $z\ne 1$. And the limit as $z$ approaches, say, $-1$ from the right $$\lim_{z\to -1^+} f(z)=\frac{1}{2};$$ the function is continuous there even though the series fails to converge. You need to understand this simple example first; it has all the essential difficulties.