Timeline for $\lim_{k\to\infty}{n\choose k}2^{1-{k\choose2}}$, where $n=\max\{n\in\mathbb{N}:{n\choose k}2^{1-{k\choose2}}<1\}$
Current License: CC BY-SA 4.0
5 events
when toggle format | what | by | license | comment | |
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Jun 29, 2018 at 22:22 | vote | accept | xFioraMstr18 | ||
Jun 29, 2018 at 20:58 | comment | added | Robert Israel | Let $f(n,k) = {n \choose k} 2^{1-{k \choose 2}}$. The point is that by definition, $$ f(N(k),k) < 1 \le f(N(k)+1,k) = \frac{N(k)+1}{N(k)+1-k} f(N(k),k)$$ so if $N(k)/k \to \infty$, $0 < 1 - f(N(k),k) \le \frac{k}{N(k)+1-k}f(N(k),k) \to 0$. | |
Jun 29, 2018 at 20:41 | comment | added | xFioraMstr18 | We have $k^3/k=k^2\to\infty$ but ${k^3\choose k}2^{1-{k\choose2}}\to0\neq1$, if I'm not mistaken; the condition doesn't seem to suffice for this case. | |
Jun 29, 2018 at 18:03 | comment | added | xFioraMstr18 | Okay, why does $N(k)/k\to\infty$ suffice? | |
Jun 29, 2018 at 16:23 | history | answered | Robert Israel | CC BY-SA 4.0 |