Is there a power series representation of $\frac{1}{\zeta(s)}$? I've seen power series representations for $\zeta(s)$, $\textit{e.g.}$
\begin{align*}
\zeta(s)=\frac{1}{s-1}+\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k!}\gamma_{k}(s-1)^{k}
\end{align*}
where $\gamma_{k}$ are the Stieltjes' constants.
Is there a reason for not having a power series representation for
$\frac{1}{\zeta(s)}$?
Thanks.
 A: Here is one I made for you $$(s-1)-\gamma  (s-1)^2+(s-1)^3 \left(\gamma _1+\gamma ^2\right)+(s-1)^4 \left(-2
   \gamma  \gamma _1-\frac{\gamma _2}{2}-\gamma ^3\right)+(s-1)^5 \left(3 \gamma ^2
   \gamma _1+\left(\gamma _1\right){}^2+\gamma  \gamma _2+\frac{\gamma _3}{6}+\gamma
   ^4\right)+\frac{1}{24} (s-1)^6 \left(-96 \gamma ^3 \gamma _1-72 \gamma 
   \left(\gamma _1\right){}^2-36 \gamma ^2 \gamma _2-24 \gamma _1 \gamma _2-8 \gamma 
   \gamma _3-\gamma _4-24 \gamma ^5\right)+O\left((s-1)^7\right)$$  where $\gamma_{k}$ are the Stieltjes' constants and $\gamma$ Euler' constant.  
You can easily build it by identification writing $$\frac{1}{\zeta(s)}=\sum _{i=0}^{\infty } a_i (s-1)^i$$ and then $$\left[\sum _{i=0}^{\infty } a_i (s-1)^i\right]\left[\frac{1}{s-1}+\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k!}\gamma_{k}(s-1)^{k}\right]-1=0$$ Canceling as many terms as desired, the previous expression is obtained up to any order (coefficient $a_n$ is obtained from term $(s-1)^{n-1}$ from a simple linear equation)
Added later
Thanks to you and thanks to Gottfried Helms's comment, I found in Wikipedia that the reciprocal of the zeta function may be expressed as a Dirichlet series over the Möbius function,
$$\frac{1}{\zeta(s)}=\sum _{k=1}^{\infty} \mu (k) k^{-s}$$ where $\mu(k)$ is the Möbius function (http://en.wikipedia.org/wiki/Riemann_zeta_function). I hate when I discover the wheel again !!
