Timeline for Abscissa of convergence of transformed Dirichlet series
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 16, 2021 at 10:16 | comment | added | reuns | Under some conjectures on the simplicity of the zeros and other stuffs, for say $a< -10$ then $\sum_n \lambda(n) n^{-s} (\log n)^a$ converges on the boundary of the domain of convergence while $\sum_n \lambda(n) n^{-s}$ doesn't. | |
| May 16, 2021 at 10:13 | comment | added | reuns | It is obvious that the abscissa of convergence of $\sum_n \lambda(n)n^{-s}$ and $\sum_n \lambda(n)n^{-s} (\log n)^a (\log \log n)^b (\log \log \log n)^c$ are the same from a partial summation, so that $\sum_n\lambda(n) n^{-s}(\log n)^a (\log \log n)^b (\log \log \log n)^c$ diverges for $\Re(s) < 1/2$. Then $\sum_n \lambda(n) n^{-0.9}$ "really seems to converge" because the RH is true up to $|\Im(s)|< 10^{15}$. | |
| May 16, 2021 at 1:49 | comment | added | Vincent Granville | @Reuns: what is obvious? The answer to my question (agree with you, though I've never seen it stated as a theorem in the literature)) or the fact that $F(0.9)$ converges if $f(k)=\lambda(k)$? (I don't think the latter one is obvious). | |
| May 16, 2021 at 1:07 | comment | added | reuns | This is obvious from a partial summation. | |
| May 15, 2021 at 19:22 | history | edited | Vincent Granville | CC BY-SA 4.0 |
added 1 character in body
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| May 15, 2021 at 19:19 | comment | added | Vincent Granville | If $f(k)=\lambda(k)$ it is conjectured that $\sigma_c(F)=\frac{1}{2}$, and that is equivalent to the Riemann hypothesis. If $f(k)=1/\sqrt{k}$ it is easy to show that $\sigma_c(F)=\frac{1}{2}$, and that $\sigma_c(F^*)=\frac{1}{2}$ both when $\alpha=1$ or $\alpha=-1$. | |
| May 15, 2021 at 17:55 | history | asked | Vincent Granville | CC BY-SA 4.0 |