Timeline for What is $\lim_{n\to\infty} \displaystyle \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\left(4^{-k}\binom{2k}{k}\right)^{\frac{2n}{\log_2{n}}}\,?$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2014 at 11:26 | history | edited | Simd | CC BY-SA 3.0 |
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| Apr 15, 2014 at 21:07 | vote | accept | Simd | ||
| Apr 15, 2014 at 18:55 | comment | added | Greg Martin | Given S. Carnahan's answer, it's clear that this really is a borderline case that's extremely interesting. Kudos for posing an insightful problem! | |
| Apr 13, 2014 at 9:44 | answer | added | S. Carnahan♦ | timeline score: 10 | |
| Apr 13, 2014 at 6:20 | comment | added | Simd | @user48365 That's interesting. Mathematica can't handle it. Can you paste the Maple code you used? If you change the $2$ in the exponent to $3$ does it then converge? | |
| Apr 12, 2014 at 20:58 | comment | added | user48365 | It can be show that in Maple soft, your limit is unbounded. | |
| Apr 12, 2014 at 19:08 | comment | added | Greg Martin | I observe that the summand seems to peak around $k \sim n/(\log_2 n\cdot \log\log n)$, so working out how big it is there would be useful. | |
| Apr 12, 2014 at 19:06 | comment | added | Simd | @GregMartin Very much the former. The constant factor seems to make a difference as to whether it converges or not. | |
| Apr 12, 2014 at 18:59 | comment | added | Greg Martin | Do you mean $\log_2n = (\log n)/(\log 2)$ or $\log_2 n = \log\log n$? | |
| Apr 12, 2014 at 17:04 | review | First posts | |||
| Apr 12, 2014 at 17:06 | |||||
| Apr 12, 2014 at 16:46 | history | asked | Simd | CC BY-SA 3.0 |