Editing Riemann zeta function (section)

===Dirichlet series===
An extension of the area of convergence can be obtained by rearranging the original series.<ref name="Knopp">{{cite book|first=Konrad|last=Knopp|title=Theory of Functions, Part Two|url=https://archive.org/details/in.ernet.dli.2015.212186|date=1947|pages=[https://archive.org/details/in.ernet.dli.2015.212186/page/n57/mode/2up 51–55]|publisher=New York, Dover publications}}</ref> The series
:<math>\zeta(s)=\frac{1}{s-1}\sum_{n=1}^\infty \left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)</math>
converges for {{math|Re(''s'') > 0}}, while
:<math>\zeta(s) =\frac{1}{s-1}\sum_{n=1}^\infty\frac{n(n+1)}{2}\left(\frac{2n+3+s}{(n+1)^{s+2}}-\frac{2n-1-s}{n^{s+2}}\right)</math>
converge even for {{math|Re(''s'') > −1}}. In this way, the area of convergence can be extended to {{math|Re(''s'') > −''k''}} for any negative integer {{math|−''k''}}.

The recurrence connection is clearly visible from the expression valid for {{math|Re(''s'') > −2}} enabling further expansion by integration by parts.

:<math>\begin{aligned}
\zeta(s)= & 1+\frac{1}{s-1}-\frac{s}{2 !}[\zeta(s+1)-1] \\
- & \frac{s(s+1)}{3 !}[\zeta(s+2)-1] \\
& -\frac{s(s+1)(s+2)}{3 !} \sum_{n=1}^{\infty} \int_0^1 \frac{t^3 d t}{(n+t)^{s+3}}
\end{aligned}</math>