Looking for series representations of Riemann's zeta function valid for $\sigma<1$.

Question

I've been looking for series representations of the Riemann's zeta function $\zeta(s)$ valid for $\sigma< 1$, with $s=\sigma + t i \in \mathbb{C}$.

I'm interested, preferably, in series representation, something like

$$ \zeta(s)= ?\sum? $$

Here is an exemple of wat I'm talking about but valid for $\sigma<0$. $$ \zeta(s)=\Gamma(1-s)\left(\sum_{k=1}^{\infty}\frac{1}{(2ki\pi)^{1-s}}+\frac{1}{(-2ki\pi)^{1-s}} \right) $$ This one is from the book Special Functions, An Introduction to the Classical Functions of Mathematical Physics by Nico M. Temme pag.58.

Please leave a reference if you post something.

Thanks.

Answer 1

How about the alternating zeta function representation

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

valid for $\Re(s)>0$.

There's also the less explicit Laurant expansion around the pole $s=1$, in terms of Stieltjes constants.

Finally, there's

$$\zeta(s)=\frac{1}{1-2^{1-s}}\sum_{n\geq 0}\frac{1}{2^{n+1}}\sum_{k= 0}^n(-1)^k\binom{n}{k}(k+1)^{-s},$$

conjectured by Knopp, and proven by Hasse, which is convergent everywhere except $s=1$.

Answer 2

What about the following:

$$\eta(s):=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=1-\frac1{2^s}+\frac1{3^s}-\ldots$$

Now

$$\begin{align*}&\bullet \zeta(s)=\sum_{n=1}^\infty\frac1{n^s}=1+\frac1{2^s}+\frac1{3^s}+\ldots\\{}\\&\bullet2^{1-s}\zeta(s)=\sum_{n=1}^\infty\frac2{(2n)^s}=\frac2{2^s}+\frac2{4^s}+\frac1{6^s}+\ldots\end{align*}$$

Rest first equation from second one above:

$$\left(1-2^{1-s}\right)\zeta(s)=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}=\ldots=\eta(s)$$

and we have the functional equation

$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$

Since $\;\eta(s)\;$ converges for Re$\,(s)>0\;$, if we don't have issues with the multiplying factor above then we've extended the zeta function to $\;0<\,$Re$\,(s)<1$...and we don't since all those are removable singularities (why?)