Timeline for What can one say about $\sum\limits_{i=1}^\infty \frac{1}{p_{i+1}^2-p_i^2}$?
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Apr 7, 2021 at 20:00 | vote | accept | Yaakov Baruch | ||
Apr 7, 2021 at 19:59 | comment | added | Yaakov Baruch | So even with effective error terms there doesn't seem to be much one can do to reign in the oscillations besides computing an outrageous number of terms. I tried printing only for the largest prime before each multiple of 5040 (hoping that this smooths the prime gaps some), up to 1e7, then picking $c$ in $\sum \frac{1}{ p_{i+1}^2 - p_i^2} + \frac{ \log \log p_{n+1} + c}{ 2\log p_{n+1} } $ to get a curve that looks like it reaches a flat plateau (not exactly a sound mathematical reason) and to guess a limit not far from 0.6365. I hope someone with superior computing power will take notice... | |
Apr 7, 2021 at 15:23 | comment | added | Zhou | @Yaakov Baruch, when we sum to $X$, the remainder of the error is $O(\log\log X /\log X)$, so the convergence rate of the series is very slow. In addition, we can give an effective error term, that is to say, the bound of the implied constant of the above $O$-term. However, this requires more numerical upper bound results of the Hardy--Littlewood prime pair conjecture and which will mainly come from the sieve method. | |
Apr 7, 2021 at 14:13 | comment | added | Yaakov Baruch | Is either the conditional or unconditional asymptotic estimate effective. In other words can one use it to derive an upper bound to $\sum_{i=1}^\infty \frac{1}{p_{i+1}^2-p_i^2}$, which converges to the limit and which combined with the trivial lower bound (or hopefully, with a better lower bound) could be used to compute the sum to within any accuracy (unpractical as that may be)? | |
Apr 7, 2021 at 9:15 | history | edited | GH from MO | CC BY-SA 4.0 |
edited for grammar and spelling
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Apr 7, 2021 at 1:01 | history | edited | Zhou | CC BY-SA 4.0 |
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Apr 7, 2021 at 0:51 | history | edited | Zhou | CC BY-SA 4.0 |
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Apr 7, 2021 at 0:45 | history | edited | Zhou | CC BY-SA 4.0 |
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Apr 7, 2021 at 0:40 | history | answered | Zhou | CC BY-SA 4.0 |