4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $Re(s) > 0$.

The second formula follows the same idea.

$\endgroup$
  • $\begingroup$ brilliant answer. But how did you imagine to get the integrals for the sums, and can we analytically continue the zeta function by this method to the whole plane? $\endgroup$ – vidyarthi Dec 25 '16 at 16:24
  • $\begingroup$ @vidyarthi I don't see what you mean. And this method is very useful in the context of Dirichlet series. $\endgroup$ – reuns Dec 25 '16 at 22:12

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.