Timeline for Conjectured relation between alternating Prime zeta series and Riemann zeta
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 29, 2014 at 11:56 | vote | accept | joro | ||
| Sep 29, 2014 at 11:46 | comment | added | joro | No, I don't have reason to expect this. | |
| Sep 29, 2014 at 10:58 | comment | added | Vladimir Dotsenko | @joro: do you have a reason to expect that? if so, it might make sense to post it as a separate question. | |
| Sep 29, 2014 at 8:50 | comment | added | joro | Thank you. Can you convert the series to integral? Possibly for fixed $n$ or dropping (-1)^k? | |
| Sep 28, 2014 at 11:12 | comment | added | Vladimir Dotsenko | @joro: of course, since it multiplies the whole sum by $-1$... | |
| Sep 28, 2014 at 11:08 | comment | added | joro | Your suggestion to replace (-1)^k with (-1)^(k-1) leads to very similar result. | |
| Sep 27, 2014 at 16:34 | comment | added | Vladimir Dotsenko | @joro: yes of course. The first equality is the definition of the prime zeta, the second is the Taylor series of logarithm, the third is basic algebra, the fourth is Euler product for zeta. | |
| Sep 27, 2014 at 16:13 | comment | added | joro | In the second sum and later sums, $\sum_{k,p}$ do you assume $p$ is prime? | |
| Sep 27, 2014 at 14:51 | history | edited | Vladimir Dotsenko | CC BY-SA 3.0 |
corrected the formulas (there was a minor sign issue)
|
| Sep 27, 2014 at 14:50 | comment | added | Vladimir Dotsenko | @joro: yes you are right, it's my mistake with signs, not yours :) | |
| Sep 27, 2014 at 14:08 | comment | added | joro | I mean $(-1)^k$. Thank you. Indeed for n=8 I need additional numerator :-). | |
| Sep 27, 2014 at 14:03 | history | answered | Vladimir Dotsenko | CC BY-SA 3.0 |