Riemann Zeta Function On Line Re(s)=1

Question

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?

Answer 1

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.