The following series for Riemann Zeta function converges for $\Re s>0$ $$\zeta(s)=\frac{1}{s-1}\sum_{n=1}^\infty(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}).$$

See for example this site or Wikipedia [1].

If there are no mistakes in my computations and the following assertions are rigth I can write easily that for a fixed zero of Riemann zeta function $s=\frac{1}{2}+t$, where $t>0$ is a real number:

$$\sum_{n=1}^{\infty}\left(\frac{n}{\sqrt{n+1}}\cos(t\log (n+1))-\frac{(n-\frac{1}{2})}{\sqrt{n}}\cos(t\log n)+\frac{t}{\sqrt{n}}\sin(t\log n)\right)=0,$$ the path is easy as I've said, first we write $m^s$ for $s=\sigma+it$, where $\sigma=\Re s$ (in our case $\frac{1}{2}$), the real part of $s$, as $m^\sigma \cdot e^{it\log m}$, after we multiply and divide by the conjugate $e^{-it\log m}$, and thirdly (I say if there are no mistakes and if it is possible, I say there is convergence for take the real part) we take the real part of

$$\sum_{n=1}^{\infty}\frac{n}{\sqrt{n+1}}\left(\cos(t\log (n+1))-i\sin(t\log (n+1))\right)$$ $$\qquad\qquad\qquad\qquad\qquad-\frac{(n-\frac{1}{2}-it)}{\sqrt{n}}\left(\cos(t\log n)-i\sin(t\log n)\right)=0.$$

My question is about

Question. If it is possible a proof verification of previous assertions in the sense if it is possible to take the real part to prove that previous series (in which we've applied real part) converges, for a fixed zero $s=\frac{1}{2}+it$, to $0$. Is know of you can write it some condition or an inequality that relates the coefficients $\frac{n}{\sqrt{n+1}}$, $-\frac{n-\frac{1}{2}}{\sqrt{n}}$ and $\frac{t}{\sqrt{n}}$ providing us this convergence to $0$*? If you need Riemann hypothesis in your answer and in fact it is neccesary for my question was well posed, you can assume this. Thanks in advance and sorry for my english.

*Remark to previous question: I say this, (first because by Godfrey's statement, see [1], it is known that on critical line there are infinitely many zeroes of Riemann zeta function, thus you can assume that $t$ is fixed but arbitrarily large, thus the conjectural condition or inequality it seems feasible) because, for example if the first and second of these coefficients were constants in its numerators then I would say by contradiction that Riemann hypothesis is false, since $\cos$ and $\sin$ are bounded and in the coefficient $t/\sqrt{n}$, $t$ is arbitraly large and we have not the exclaimed convergence to 0. Truly our first two coefficients are not constants in its numerators, then I believe that this condition or inequality should exists, at least assuming Riemann hypothesis as hypothesis to work.


[1] Wikipedia, Riemann zeta function , Representations https://en.wikipedia.org/wiki/Riemann_zeta_function

  • $\begingroup$ I hope that previous is a well question to this Mathematics Stack Exchange, I don't know if previous is in literature, and I believe that it is reasonable and possible give an useful answer in this site. My goal in this site is learn with easy facts and computations. In other case, if it is a bad question or are mistakes, please tell me, I can delete it. Thanks. $\endgroup$ – user243301 Nov 20 '15 at 22:46
  • $\begingroup$ If there is an answer, it is possible that I will delete my last Remark to provide to this site the best useful answer/question, since my remark is specualtive in the way that I've tried explain my question. What I want to express in my question is: what condition in coefficients provide us the convergence to $0$? I believe that there is such condition. Thanks. $\endgroup$ – user243301 Nov 21 '15 at 14:52

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