Riemann Zeta Function at $z=1+i \cdot t$ First, we define the Riemann Zeta functon as a infinite sum:
$$\zeta(z)=\sum_{n=1}^{\infty}\frac{1}{n^z}, \ Re(z)>1$$
This is an analytic function in this domain. We can not extend this function with this summation because we have
$$\left|\frac{1}{n^z}\right|\leq \left|\frac{1}{n^{Re(z)}}\right|$$
and then by comparision the entire sum converges when $Re(z)>1$.
We also know that we can extend $\zeta$ via holomorphic continuation to a meromorphic function with a simple pole in $z=1$.
My question is, what about the values of the extended $\zeta$ function in $1+it$, where $t\in\mathbb{R}$? I mean, if you take the limit of the summation by the right way, you have this
$$\lim_{z\rightarrow (1+it)^+}\sum_{n=1}^{\infty}\frac{1}{n^z}$$
and this converges to $\zeta(1+it)$ because is an analytic continuiation, hence continuous, right? But the sum goes to infinity, or am I wrong?
Thank you very much!
 A: You seem to think that because $\lim_{\sigma\to1^+}\sum_n n^{-(\sigma+it)}$ exists (which is true by the reasons you mention), it should be equal to $\sum_n n^{-(1+it)}$. But that's not true unfortunately. It's the whole shebang about when we can interchange limits and sums. As a simpler analogous example:
$$
f(x) := \frac{1}{1-x} = \sum_{n\ge 0}x^n
$$
for $|x|<1$. Here $f$ is continuous at $x=-1$ with $f(-1)=\frac12$, but the series is $1-1+1-1+\cdots$ and diverges. It's just how it is.
Note also that the series diverges without blowing up to infinity, just as the $\zeta$ funciton series at $1+it$.
A: For $s\ne 1$
$$\lim_{N\to \infty} \sum_{n= 1}^N n^{-s}=\lim_{N\to \infty}\left( \int_1^{N+1} x^{-s}dx+\sum_{n= 1}^N (n^{-s}-\int_n^{n+1} x^{-s}dx)\right)$$
$$ = \lim_{N\to \infty}\left(\frac{1-(N+1)^{1-s}}{s-1}+ \sum_{n= 1}^N \int_n^{n+1}\int_n^x st^{-s-1}dtdx\right)$$

*

*$\lim_{N\to \infty}\sum_{n= 1}^N \int_n^{n+1}\int_n^x st^{-s-1}dtdx$ converges for $\Re(s) > 0$


*$\lim_{N\to \infty}\frac{1-(N+1)^{1-s}}{s-1}$ converges iff $\Re(s) > 1$, for $\Re(s)=1,s\ne 1$ it oscillates staying bounded.
Whence $\lim_{N\to \infty} \sum_{n= 1}^N n^{-s}$ converges iff $\Re(s) > 1$. The same holds for the real and imaginary part.
The analytic continuation to $\Re(s)>0,s\ne 1$ is $\frac1{s-1}+\sum_{n= 1}^\infty \int_n^{n+1}\int_n^x st^{-s-1}dtdx$.
A: Intuitive reasoning: Given a complex number $z=1+\varepsilon+it$ for $0<\varepsilon$ and $t\neq0$, the partial sums
$$
\sum_{n=1}^k\frac1{n^z}
$$
form a spiral-like pattern, spiralling inwards to $\zeta(z)$. There is, a priori, no reason why the limit of the partial sums should go to infinity as $\varepsilon\to0$. And in fact, it turns out that doesn't happen. The spiral just spirals in slower and slower, and the center it spirals towards doesn't fly off to infinity. At $1+it$ the partial sums all lie along a circle (this circle is larger the smaller $|t|$ is)
