Timeline for Uniform upper bound for the sum over primes $\sum_{p \leq x} p^{-1+\varepsilon}$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2015 at 23:21 | comment | added | Christian Remling | Definitely I'm in great shape with the OP's extra specification which makes $\log\log x/\epsilon$ small. | |
| Apr 23, 2015 at 23:10 | comment | added | Christian Remling | @FanZheng: Yes, I guess that's what I show: $\limsup S_{\epsilon}(x)x^{-\epsilon}\log x\le C/\epsilon$, and this $C$ doesn't depend on anything. | |
| Apr 23, 2015 at 21:21 | comment | added | Fan Zheng | Maybe I'm confused, but that's true only in some sense of the two "sufficiently" above. In other words, does the "sufficient largeness" of $x$ depend on $\epsilon$? | |
| Apr 23, 2015 at 21:17 | comment | added | Christian Remling | @FanZheng: It sure is; $\log\log x\le (\epsilon/2)\log x$, say. | |
| Apr 23, 2015 at 21:14 | comment | added | Fan Zheng | But is the estimate uniform in $\epsilon$, e.g., in the second part $\log t\ge \epsilon\log x-\log\log x$. Why is it $\gg \epsilon\log x$ with a uniform constant? | |
| Apr 23, 2015 at 20:58 | history | answered | Christian Remling | CC BY-SA 3.0 |