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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?

Asier Calbet
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  • 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 04 '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 04 '14 at 10:39
  • Did you mean "from the left side" in your comment? – Mikhail Katz Aug 04 '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 04 '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 04 '14 at 10:56
  • Could you please define uniformly convergent for me? – Asier Calbet Aug 04 '14 at 10:58
  • http://en.wikipedia.org/wiki/Uniform_convergence – Mikhail Katz Aug 04 '14 at 11:00
  • So an analytic function, despite often being called "nice" functions, need not be uniformly convergent? – Asier Calbet Aug 04 '14 at 11:04

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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.

stopple
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  • Thanks for the answer, curiously enough, weeks after I wrote this question I thought of EXACTLY this example to answer my own question. Thanks anyways, I appreciate it. – Asier Calbet Sep 05 '14 at 15:04
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    The geometric series is the answer to quite a lot of questions. – stopple Sep 05 '14 at 19:03
  • I am lost as how does this answers the question. That $1/(1-z)$ has no problems at $z=-1$ is easy to see, but what happens to the series when $z\to-1^+$? I tried to compute some values of the limit and it looks as if it oscillates but is bounded. Is this typical for series that partly represent a holomorphic function? I am not sure if this rewards a new question. Rather I think it would be great to extend the answer a bit. – Harald May 24 '20 at 06:19