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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