# Derivation of a series expansion for Riemann Zeta function

Though there is a vast amount of literature on Riemann's Zeta Function, but, I was struck by this formula of the series, which is given in Wikipedia. Even after seeing several tracts from this site as well as others, I am unable to find its derivation. If it is a duplicate, or the question is asked before, please let me know. The question is how does the below two series, given in the link above, be derived? $$\zeta(s)=\frac{1}{s-1}\sum_{n=1}^{\infty}\left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)\forall \Re(s)>0$$ and $$\zeta(s)=\frac{1}{s-1}\sum_{n=1}^{\infty}\frac{n(n+1)}{2}\left(\frac{2n+3+s}{(n+1)^{s+2}}-\frac{2n-1-s}{n^{s+2}}\right)\forall \Re(s)>-1$$

I once got a comment that these formulae are untrue. Are their claims right? But, since it is there in Wkipedia, it deserves some crucial examination. The series are instructive in that,if they are true, they give us analytic continuation without the use of Integration. I guess it is by using the Mittag-Leffler Theorem as the reference in the article points to Knopp's Theory of Functions book, if which I have no copy but am unsure. Thanks beforehand

For $Re(s) > 2$ : $$\sum_{n=1}^{\infty}\left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)= \sum_{n=1}^\infty\left( (n+1)^{1-s}-(n+1)^{-s}- n^{1-s}+s n^{-s} \right)$$ $$= \zeta(s-1)-1-(\zeta(s)-1)-\zeta(s-1)+s \zeta(s) = (s-1)\zeta(s)$$ since $\frac{n}{(n+1)^s}-\frac{n-s}{n^s} = sn\int_n^{n+1} (n^{-s-1}-x^{-s-1})dx$ $=s n\int_n^{n+1} \int_n^x (s+1)t^{-s-2}dtdx= \mathcal{O}(n^{-s-1})$
$\sum_{n=1}^{\infty}\left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right)$ converges and is analytic for $Re(s) > 0$, and by analytic continuation $\sum_{n=1}^{\infty}\left(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\right) = (s-1) \zeta(s)$ stays true for $Re(s) > 0$.