Timeline for Sum f(p) over all primes convergent with sum f(n) over all natural numbers divergent?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 6, 2010 at 23:02 | answer | added | Pietro Majer | timeline score: 5 | |
| Jul 6, 2010 at 22:05 | comment | added | Michael Lugo | Qiaochu, I think you mean $f(n) = 1/(n \log n)$. | |
| Jul 6, 2010 at 21:40 | vote | accept | Andreas Rüdinger | ||
| Jul 6, 2010 at 21:28 | answer | added | Fedor Petrov | timeline score: 11 | |
| Jul 6, 2010 at 21:24 | comment | added | David E Speyer | Switch converges and diverges in my comment. | |
| Jul 6, 2010 at 21:22 | comment | added | Qiaochu Yuan | f(n) = n log n has that property, doesn't it? | |
| Jul 6, 2010 at 21:15 | comment | added | David E Speyer | Or, easier to think about, a function $f(t)$ such that $\int f(t) dt$ converges and $\int f(t) dt/\log t$ diverges. | |
| Jul 6, 2010 at 21:12 | comment | added | Noah Snyder | The simple approach to this problem is to replace \sum_p f(p) with the roughly equivalent \sum_n f(n log n) since the nth prime is roughly n log n. Once you've found a monotonic function where \sum_n f(n log n) converges but \sum_n f(n) diverges then it probably won't be too hard to use the prime number theorem to answer your question. | |
| Jul 6, 2010 at 21:01 | history | asked | Andreas Rüdinger | CC BY-SA 2.5 |