# Upper bound for real part of Riemann Zeta function zeros

Hi,

I have been reading about Riemann Zeta function $\zeta(s)$ and have been thinking about it for some time. I did some calculations, and reached a conclusion where $\Re(\rho) \le \log_2(3) - 1$ as $\Im(\rho) \to \infty$ where $\rho$'s are the roots of Riemann Zeta function in the critical strip. Anyways, I know its not the place to discuss claimed proofs and similar stuff, but just to give a background of where I am coming from. So straight to the question.

Is there any similar result regarding upper bound ($< 1$) for the real part of the zeros zeta function as their imaginary parts tend to infinity?

Thanks

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All results on zero-free regions have a strip with width tending to 0 as the imaginary part increases. –  Charles Matthews Oct 5 '10 at 8:40
All known results on zero free regions are of the form $\zeta(\sigma+it) \neq 0$ when $\sigma > 1 - f(t)$, for some function $f$ with $\lim_{t \to \infty} f(t) =0$. A bound of the form you describe would be major progress. –  David Speyer Oct 5 '10 at 10:49
Honestly, you almostly certainly have an error. Your past questions have involved a number of elementary errors about analytic number theory and you are working on a problem which is notorious for drawing mathematicians into false proofs. In addition, if you are who I think you are, than 3 of your 6 papers on the arXiv have been withdrawn. (I am glad that you do withdraw errors!) Based on the little I know about you, you really need to find an advisor or mentor who can teach you how to do careful work in analytic number theory. –  David Speyer Oct 5 '10 at 11:00
If the proof is incorrect, what is the first equation in the paper that is incorrect? –  user13672 Mar 15 '11 at 19:16
You could look for yourself. I did, and the limit $\tau \to \infty$ at the beginning of Section 4 looks pretty dubious. –  Emerton Mar 15 '11 at 19:31

I was searching for papers having zero free regions... and most of them gave results like $\Re(\rho) > 1 - \beta$. Are you sure there are no such results? (I'll wait before accepting your answer) –  Roupam Ghosh Oct 5 '10 at 9:36
EDIT: by "no such results" I mean no results with $\Re(\rho) < c < 1$. Sorry for the typo –  Roupam Ghosh Oct 5 '10 at 9:39