Let $S$ be a graded ring, do we have $\mathcal O_{\operatorname{Proj}S}(\operatorname{Proj}S)=S_0$?
If $\operatorname{Re}(s)>1$, then $p^{\operatorname{Re}(s)}>2$ for any prime $p$, $p^{\operatorname{Re}(s)}-1>\frac{1}{2}p^{\operatorname{Re}(s)}$,
\begin{eqnarray} \zeta(s)=\sum\limits_{n=1}^\infty\frac{1}{n^s}&=&\prod\limits_p\frac{1}{1-p^{-s}}\\ &=&\prod\limits_p\frac{p^s}{p^s-1}\\ &=&\prod\limits_p(1+\frac{1}{p^s-1}) \end{eqnarray}
\begin{eqnarray} \sum\limits_p|\frac{1}{p^s-1}| &=& \sum\limits_p\frac{1}{|p^s-1|}\\ &\le & \sum\limits_p\frac{1}{|p^s|-1}\\ &=&\sum\limits_p\frac{1}{p^{\operatorname{Re}(s)}-1}\\ &<& \sum\limits_p\frac{1}{\frac{1}{2}p^{\operatorname{Re}(s)}}\\ &<&2\sum\limits_{n=1}^\infty\frac{1}{n^{\operatorname{Re}(s)}}\\ &=&2\zeta(\operatorname{Re}(s))\\ &<&\infty \end{eqnarray}
$\forall s$ with $\operatorname{Re}(s)>1$, for any prime $p$, $\frac{1}{1-p^{-s}}\neq 0$, so $\zeta(s)=\sum\limits_{n=1}^\infty\frac{1}{n^s}&=&\prod\limits_p\frac{1}{1-p^{-s}}\neq 0$,
that is, $\zeta(s)$ does not vanish when $\operatorname{Re}(s)>1$.
This seems to be true if $S$ is a polynomial ring. This follows from Hartshorne Prop II.5.13.
It also states that if $S$ is a graded ring that is not a polynomial ring, then this is not true. Exercise 5.14 provides a counterexample.