Timeline for Determining the asymptotic behavior of a series
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2011 at 0:57 | vote | accept | Erwin | ||
| Apr 14, 2011 at 0:57 | |||||
| Apr 14, 2011 at 0:56 | vote | accept | Erwin | ||
| Apr 14, 2011 at 0:56 | |||||
| Apr 13, 2011 at 1:24 | comment | added | Peter Humphries | Oops, sorry, you are right. My method doesn't seem to be able to get that $O(1 - \mu)$ either. | |
| Apr 13, 2011 at 1:19 | comment | added | GH from MO | @Peter: Your bounds $\mu/(1-\mu)$ and $1/(1-\mu)$ differ by $1$, i.e. not by $O(1-\mu)$. However, I might have made a mistake in my quick notes, i.e. I would not swear that $O(1-\mu)$ indeed follows from our ideas. Note that the limit provably fails to exist in some cases, see the "EDIT" part at the end of my response. BTW I found your approach rather elegant! | |
| Apr 13, 2011 at 1:09 | comment | added | Peter Humphries | @GH: Yes, I meant to show bounds on the $\liminf$ and $\limsup$; I wasn't actually claiming that the limit itself exists. The bounds I do prove though show that the bounds are indeed $O(1 - \mu)$ of each other, and that for $\mu$ close to $1$ the $\liminf$ looks more and more like the $\limsup$, but this isn't too useful a heuristic as my bounds (and yours) show that both the $\liminf$ and the $\limsup$ must blow up at $\mu = 1$. | |
| Apr 13, 2011 at 1:03 | history | edited | Peter Humphries | CC BY-SA 3.0 |
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| Apr 12, 2011 at 22:25 | comment | added | GH from MO | @Erwin: From the two responses you can see how $nf_n(t)$ oscillates between two positive constants depending on $\mu$. Combining my ideas with that of Peter I think I can prove that the two constants are $O(1-\mu)$ apart from each other, i.e. for $\mu$ very close to $1$ the limit "almost exits". I can also believe that for generic $\mu$'s the limit exists, but this would be tricky to prove. For special $\mu$'s (e.g. the reciprocal of a square number) the limit fails to exist for simple diophantine reasons. See the "EDIT" section in my response. | |
| Apr 12, 2011 at 21:59 | comment | added | Erwin | Thanks very much for your responses. I had tried summation by parts but did not write the second summation as an integral, which leads to the nice estimates of Peter Humphries. I had also done some analysis in the spirit of GH though not as deep, that's how I came to believe that $O(1/n)$ was the right rate of decay. I have done other things and have some ideas that I will be posting tomorrow or today if I have a chance. If the limit $\lim_{n\to\infty}nf_n(t)$ does not exist, then the natural question is how does $nf_n(t)$ behave? Thanks again one more time. | |
| Apr 12, 2011 at 15:20 | comment | added | GH from MO | More precisely, you give bounds for the $\liminf$ and the $\limsup$ that are similar to mine (in fact better than mine, but my argument can also be made effective in a similar fashion). | |
| Apr 12, 2011 at 15:17 | comment | added | GH from MO | @Peter: I have not followed every step in your argument, but you don't seem to prove that $\lim_{n\to\infty} nf_n(t)$ exists. All you prove is that $ntf_n(t)$ lies between two positive constants depending on $\mu$. This is in agreement with my finding above. However, I believe that $\lim_{n\to\infty} nf_n(t)$ does not exist: based on the diophantine properties of $n$ you can produce different limits along subsequences. | |
| Apr 12, 2011 at 10:31 | history | edited | Peter Humphries | CC BY-SA 3.0 |
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| Apr 12, 2011 at 9:27 | comment | added | Fabian | @Peter Humphries: a guess there must be a sign mistake. The original sum was positive and your estimates are negative ... | |
| Apr 12, 2011 at 9:20 | history | answered | Peter Humphries | CC BY-SA 3.0 |