How does $\zeta(i\pi)$ converge?

Question

$$\zeta(i\pi) = \sum_{r=1}^{\infty}r^{-i\pi} = \sum_{r=1}^{\infty}e^{-i\pi \ln(r)} = \sum_{r=1}^{\infty}\operatorname{cis}(-\pi\ln(r))$$

Did I mess up somewhere in the steps above? I can't see how the last expression would converge.

Answer 1

Your mistake is thinking that $\zeta(i \pi)$ is defined by that sum.

That summation only defines $\zeta$ for complex numbers with real part bigger than one. This definition is extended to the rest of the complex plane by the process of analytic continuation.

The basic theoretical (but not practical) idea is that once you know $\zeta(z)$ for such complex numbers, you could then, say, find its Taylor expansion around the point $1.01 + 3i$. This Taylor expansion has radius of convergence larger than 0.01, and so it lets you define $\zeta(z)$ for even more numbers. And you keep going until you get everywhere in the plane.

And happily, $\zeta(z)$ is a sufficiently nice function that you can do this consistently. More annoying functions would have branch cuts and "monodromy" and stuff.

Of course, this theoretical description is only really useful for explaining that it can be done, not how to actually compute it in practice or study it fruitfully in theory. That uses tools like the functional equation for $\zeta$, or the construction of other series that sum to $\zeta$. Your wolfram alpha link gives a few of those.