What is a series expansion for $\zeta(s)$ valid iff $\operatorname{Re}(s)>\frac{1}{2}$?

Question

Let $s$ be complex and $\zeta(s)$ the Riemann Zeta function. What is a series expansion for $\zeta(s)$ valid iff $\operatorname{Re}(s)>\frac{1}{2}$ ?

I want a series expansion such that $\zeta(s)=\sum_{n}^{\infty} f(n,s)$ where the $f(n,s)$ are standard functions without irrational constants.

Answer 1

Not really an answer but perhaps close.

A simple solution without the conditions that $f(n,s)$ are standard functions though is relatively easily obtained by :

$$\zeta^*(s)=\frac 1{1-2^{1-s}}\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^s}$$

$$\zeta^{**}(s)=\sum_{n=1}^\infty \frac {1}{n^{s+\frac{1}{2}}}$$

$\zeta(s)=\dfrac{\zeta^*(s)\zeta^{**}(s)}{\zeta^{*}(s+\dfrac{1}{2})}$

and then simplifying $\zeta^*(s)\zeta^{**}(s)$ to a single sum by using a sum over the divisors.

However I would dare to call this a "nonelegant" solution.

Also I think there is another solution but not independant of RH. (using series for the multiplicative inverse of $\zeta(s)$ with conjectured domains of convergeance or such )

Bye uniqueness of analytic continuation ( but not of series representations of them ! ) I doubt that a answer to the OP is possible since that would seem like an oversimplification of deep number theoretical stuff like sum over divisors in unexpected ways...

Im speculating here so this finishes my partial answer/comments.

Answer 2

Hints:

For Re$(s)>0\,\;,\;\;s\neq 1$ :

$$\zeta(s)=\frac 1{1-2^{1-s}}\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^s}$$

You must check (1) that the infintie series there converges in the wanted domain and (2) the poles of the factor multiplying the series ad in fact removable (or "fake poles", if you will)