1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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?)

$\endgroup$
0
$\begingroup$

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 $s=1$.

$\endgroup$
  • $\begingroup$ hmmm, I asked for a representation valid for $\Re (s) < 1$ but this one is valid for $\Re (s) > 0$ $\endgroup$ – Neves Nov 27 '14 at 18:18
  • $\begingroup$ Then the linked Laurant expansion is what you're looking for. $\endgroup$ – Alex R. Nov 27 '14 at 18:19
  • $\begingroup$ No, I'm looking for something like (in the spirit of) the exemple in my question, something involving a combination of Dirichlet's series. $\endgroup$ – Neves Nov 27 '14 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.