Does $\sum_{k=1}^\infty\frac1{k^n}$ converge for $\Re(n)=1,\Im(n)\ne0$?

Question

Does $\sum_{k=1}^\infty\frac1{k^n}$ converge for $\Re(n)=1,\Im(n)\ne0$?

The ratio test is inconclusive.

It passes the term test for $\Re(n)=1$, but this is not sufficient to prove convergence.

Since we are dealing with so many complex numbers, I do not know of any convergence tests for this. I know that if it is convergent, it is conditionally convergent.

Answer 1

$$n^{-s} - \int_n^{n+1} x^{-s} dx = \int_n^{n+1} (n^{-s}-x^{-s}) dx = \int_n^{n+1} \int_n^x s y^{-s-1}dy dx = \mathcal{O}(s n^{-s-1})$$ so that $$\begin{eqnarray}\sum_{n=1}^\infty (n^{-s}-\int_n^{n+1} x^{-s} dx) &=& \lim_{N \to \infty} \sum_{n=1}^N n^{-s}- \int_1^{N+1}x^{-s}dx \\ &=& \lim_{N \to \infty} \sum_{n=1}^N n^{-s} + \frac{(N+1)^{1-s}-1}{s-1}\underset{\qquad \scriptstyle (s \ne 1)}{}\end{eqnarray}$$ converges for $Re(s) > 0$.

Since $N^{1-s}$ diverges whenever $Re(s) \le 1$, so does $\sum_{n=1}^N n^{-s}$.

Note that by analytic continuation : $$\zeta(s) = \frac{1}{s-1}+ \sum_{n=1}^\infty (n^{-s} - \int_n^{n+1} x^{-s} dx)\qquad \quad (Re(s) > 0)$$

Answer 2

In THIS ANSWER, I used the Euler-Maclaurin Summation Formula (EMSF) to show that the series $\sum_{k=1}^\infty \frac{1}{k^n}$ converges for $\text{Re}(n)>1$ and diverges for $\text{Re}(n)<1$.

Proceeding similarly, using the EMSF we can write for $n=1+iy$

$$\begin{align} \sum_{k=1}^N \frac{1}{k^{1+iy}}&=\sum_{k=1}^N \frac{e^{-iy\log(k)}}{k}\\\\ &=\frac{\sin(y\log(N))}{y}+i\frac{\cos(y\log(N))}{y}+C+O\left(\frac1N\right) \end{align}$$

for some constant $C$.

Since both $\lim_{N\to \infty}\sin(y\log(N))$ and $\lim_{N\to \infty}\cos(y\log(N))$ diverge for $y\ne 0$, the series of interest diverges likewise.