How Riemann made Z(s) to converge for s>0?

Question

I am going through the procedure that Riemann took to expand the Euler's Z-funtion. But I can not understand first part of it. We know that Z(s) just converge for s>1. How it is possible to make series converge for $0<s<1$ by just multiplying and the dividing by $(1-2/2^s)$? To give an example $$ Z(\frac{1}{2})=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+... $$ which diverges to $+\infty$. By multiplying it by $1-2/\sqrt{2}$ and re-arranging we get $$ (1-2/\sqrt{2})\cdot Z(\frac{1}{2})=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-... $$ which converges! Oddly for me, $1-2/\sqrt{2}$ is a negative number and I expect that multiplying it in $Z(\frac{1}{2})$ which diverges to $+\infty$ to give a negative result but mentioned re-arranged series is converging to a positve number!

Answer 1

The Riemann zeta function $\zeta(s)$ is defined for $\operatorname{Re}(s) > 1$ by the absolutely convergent series $\zeta(s) = 1 + 2^{-s} + 3^{-s} + \cdots$.

The series $1 + 2^{-s} + 3^{-s} + \cdots$ does not converge when $s = \frac{1}{2}$, nor does it converge when multiplied by the constant $1 - \frac{2}{\sqrt{2}}$.

Rather, there exists a unique meromorphic function $f$ from $\{ s \in \mathbb C : \operatorname{Re}(s) > 0\} \rightarrow \mathbb C \cup \{\infty\}$, such that $f(s) = \zeta(s)$ whenever $\operatorname{Re}(s) > 1$. By convention, this function $f$ is also called the Riemann zeta function and is also denoted by $\zeta$. The formula $1 + 2^{-s} + 3^{-s} + \cdots$ for $f$ no longer works when $s$ has real part less than or equal to $1$.