Timeline for Convergence of a double sum involving prime numbers
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18 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2016 at 22:52 | history | edited | GH from MO |
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| Jan 19, 2016 at 12:27 | answer | added | Fedor Petrov | timeline score: 6 | |
| Jan 19, 2016 at 4:47 | comment | added | Gerhard Paseman | Now I get it too. Fixing p, the sum of the fractional parts is bounded by 1 + (ln n - ln(p-1))/ ln p, so that the limit is 1 comes as no surprise. Sorry for my poorly thought skepticism earlier. Gerhard "Should Also Go To Sleep" Paseman, 2016.01.18 | |
| Jan 18, 2016 at 23:23 | comment | added | Brian | @alpoge Ahhh, I see. Your suggestions definitely work here. If we split the sum into $k = 0$ and $k > 0$, we have $$\epsilon_n = \frac{1}{n}\left( \sum_{p \leq n} \ln p \left\{\frac{n}{p-1}\right\} + \sum_{k > 0} \ln p \left\{\frac{n}{(p-1)p^k}\right\} \right)$$ $$\leq \frac{1}{n}\left( \sum_{p \leq n} \ln p + \frac{n\ln p}{p-1}\sum_{k > 0} \frac{1}{p^k}\right)$$ $$\sim 1+\sum_{p \leq n}\frac{\ln p}{(p-1)^2} $$ Using the same notation as in the question, we have that, for large $n$, $\epsilon_n \leq 1+C$, where $C = \lim_{n \to \infty} C_n$. | |
| Jan 18, 2016 at 23:05 | comment | added | Fedor Petrov | And well, it looks that the limit does exist, but I am sleepy now and have to recheck this. | |
| Jan 18, 2016 at 22:52 | comment | added | alpoge | Yeah, that's why I split off the k=0 term and handled it separately. | |
| Jan 18, 2016 at 22:51 | comment | added | Brian | @alpoge Gerhard Paseman may have point. Your latter fact would be true regarding that sum, but the infinite sum $\sum_{k\geq 0} \frac{1}{p^k} = \frac{p}{p-1}$ implies that our sum is $\sum_{p\leq n} \frac{p \log p}{(p-1)^2}$, which diverges as $n \to \infty$. | |
| Jan 18, 2016 at 22:36 | comment | added | Fedor Petrov | It is equivalent to the limit of $\frac{\ln n}n \sum \{\frac{n}{p-1}\}$. | |
| Jan 18, 2016 at 21:11 | comment | added | Gerhard Paseman | Brian, it is a rough idea. If I wanted to "break" your conjecture, I would try showing there is f(n) such that n+1,... n+f(n) has lots of multples of (p-1)p^k for varying p and k, enough to show divergence. I have no idea what f would break it but f=O(ln n) is where I would start. alpoge, since the fractional part behaves wierdly, I have serious doubts about your suggested approach, as the last fact seems unapplicable. Gerhard "Unapplicable Is A Word, Yes?" Paseman, 2016.01.18. | |
| Jan 18, 2016 at 20:51 | comment | added | alpoge | For a bound on the limsup, split the sum into k=0 and k>0. For the former use {x}\leq 1 and \sum_{p\leq n} \log{p} ~ n, for the latter use {x}\leq x and \sum_p \log{p}/(p(p-1)) \leq O(1). | |
| Jan 18, 2016 at 20:26 | comment | added | Brian | This attack sounds very plausible. Would you like to elaborate a bit more, or is this a rough idea at the moment? | |
| Jan 18, 2016 at 20:18 | comment | added | Gerhard Paseman | Unfortunately, the following consideration is a fly in the convergence ointment: Suppose n+1 is a multple of (p-1)p^k. Then part of your infinite sum with that p approaches k. By itself, this might not be a problem, but there is also n+2, n+3, and so on to consider. If your sum converges, you need that intervals below multiples of (p-1)p^k stay sparse enough as k grows. Gerhard "Should I Say Soup Instead?" Paseman, 2016.01.18 | |
| Jan 18, 2016 at 20:05 | comment | added | Gerhard Paseman | Oops. I missed the prime in your comment. Sorry for the confused response. I think then the sum converges, and I will see if I can help you convince others. Gerhard "Wants Confusion Replaced With Conviction" Paseman, 2016.01.18 | |
| Jan 18, 2016 at 20:03 | comment | added | Brian | The sum $$\sum_{p \leq n} \log p$$ is known as the first Chebyshev function, and it's a proven fact it is $\sim n$. | |
| Jan 18, 2016 at 19:49 | comment | added | Gerhard Paseman | Ok. If p ranged over primes, I could see sum ln p looking like n, as I think you stated. However, you have a larger quantity, so I don't know what you mean by ~ n now. Gerhard "Maybe It's A Spanish Thing?" Paseman, 2016.01.18. | |
| Jan 18, 2016 at 19:45 | comment | added | Brian | @GerhardPaseman Sorry for the being clear. $p \leq n$ is supposed to be all positive integral prime values for $p$ less than or equal to $n$. | |
| Jan 18, 2016 at 19:38 | comment | added | Gerhard Paseman | Is $p\leq n$ supposed to be positive integral $p$ at most $n$, or positive integral prime values for $p$? Gerhard "Conventions Unclear For the Unconventional" Paseman, 2016.01.18 | |
| Jan 18, 2016 at 19:04 | history | asked | Brian | CC BY-SA 3.0 |