The Riemann Zeta Function

ζ(s)=∑nn−s when that makes sense. But ζ has an "analytic continuation" to much of the rest of the complex plane.

Consider the equality:

11−z=∑n=0∞zn

The right side only converges and equals the left side when |z|<1.

But complex analysis has this awesome feature - that any function has at most one analytic continuations to the regions where it is not already defined.[*] So 11−z is the only analytic continuation of the right side for complex z.

The same is true for the ζ function. We first define it for s where the real part is greater than one. And then we find a way to continue that function for other values of s.

The heart of the extension of ζ, at least to 1/2, is that, for s>1:

(1−12s−1)∑n1ns=∑n1ns−2∑n1(2n)s=∑n(−1)n−1ns

The right side is defined for any s∈(0,1], since it is the sum of an alternating decreasing sequence. (It converges for other s with Res>0, but it isn't 100% obvious looking at it that this is true.[**])

This lets us extend ζ(s) to s∈(0,1):

ζ(s)=11−12s−1∑n(−1)n−1ns

This is equal to our original definition when Res>1.

So ζ(1/2)=11−2–√∑n(−1)n−1n−−√

Computing for M=200,000,000 terms of this sum I get:

11−2–√∑n=1M(−1)n−1n−−√≈−1.460269

This series converges very slowly.

Source of explanation: https://math.stackexchange.com/questions/1613392/why-zeta-1-2-1-4603545088