Timeline for Limit of $\sum (-1)^ {\log \log n }$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 3, 2019 at 14:10 | review | Close votes | |||
| Aug 8, 2019 at 3:05 | |||||
| Jul 26, 2019 at 20:32 | comment | added | alpoge | If you’re trying to estimate \sum z^{\omega(n)} by substituting \log \log n for \omega(n) you’re going to get wrong answers, even taking z = 2 fails. (The point is that, first off, there are huge deviations (like \sqrt{\log\log n}) from the mean, and, second off, the function z^{\omega(n)} changes by a ton even when there are small deviations.) The growth rate of \sum_{n\leq X} z^{\omega(n)} is X (\log X)^{z-1}, even for complex z, though I don’t remember the hypotheses on z off the top of my head. Look up the Selberg-Delange method. Hope that helps!! | |
| Jul 26, 2019 at 11:22 | comment | added | Gerald Edgar | Well, at least $\int_3^\infty \exp(i\pi\log\log x)\;dx$ diverges. | |
| Jul 25, 2019 at 23:48 | comment | added | Gerhard Paseman | Actually, log log x - 1 is log log (x^(1/e)), so there will be some small cancellation/deviation. Say the sum is more like (n-n^{1/e})(-1)^{log log (n/2)), or something similar. Gerhard "There With A Delicate Touch" Paseman, 2019.07.25. | |
| Jul 25, 2019 at 23:43 | comment | added | Greg Martin | The limit will not exist. $\log \log n$ grows so slowly that the sum will be overwhelmingly close to $n(-1)^{\log\log n}$. | |
| Jul 25, 2019 at 22:45 | review | Close votes | |||
| Jul 28, 2019 at 17:04 | |||||
| Jul 25, 2019 at 21:58 | history | edited | Raj Raina | CC BY-SA 4.0 |
deleted 12 characters in body; edited title
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| Jul 25, 2019 at 21:45 | history | edited | Raj Raina | CC BY-SA 4.0 |
added 442 characters in body
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| Jul 25, 2019 at 21:19 | history | edited | Raj Raina | CC BY-SA 4.0 |
edited title
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| Jul 25, 2019 at 21:11 | history | edited | Raj Raina | CC BY-SA 4.0 |
added 5 characters in body; edited title
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| Jul 25, 2019 at 21:08 | comment | added | Raj Raina | Sure, I edited the question to make it more clear | |
| Jul 25, 2019 at 20:32 | history | edited | LSpice | CC BY-SA 4.0 |
cis -> \operatorname{cis}
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| Jul 25, 2019 at 20:31 | comment | added | LSpice | Presumably 'cis' means $\exp(i\cdot) : t \mapsto \cos(t) + i\sin(t)$, so why not say that? Also, does $\lim_n \sum_n$ mean $\sum_{n = 2}^\infty$? | |
| Jul 25, 2019 at 20:26 | history | asked | Raj Raina | CC BY-SA 4.0 |