# $\lim_{k\to\infty}{n\choose k}2^{1-{k\choose2}}$, where $n=\max\{n\in\mathbb{N}:{n\choose k}2^{1-{k\choose2}}<1\}$

How do you prove $\lim_{k\to\infty,k\in\mathbb{N}}{n\choose k}2^{1-{k\choose2}}=1$, where $n=\max\{n\in\mathbb{N}:{n\choose k}2^{1-{k\choose2}}<1\}$? The expression ${n\choose k}2^{1-{k\choose2}}$ comes up in a lower bound for the Ramsey number $R(k,k)$, and I want to fill in the steps in the derivation of the asymptotics of $n$.

I have tried playing around with identities like $${m\choose k+1}2^{1-{k+1\choose2}}=\frac{m-k}{k+1}2^{-k}{m\choose k}2^{1-{k\choose2}}$$ and $${m+1\choose k+1}2^{1-{k+1\choose2}}=\frac{m+1}{m-k}{m\choose k+1}2^{1-{k+1\choose2}},$$ but I didn't get anywhere.

Spencer, Joel, Ten lectures on the probabilistic method., CBMS-NSF Regional Conference Series in Applied Mathematics. 64. Philadelphia, PA: SIAM, Society for Industrial and Applied Mathematics. vi, 88 p. (1994). ZBL0822.05060.

Let $N(k) = \max \{n \in \mathbb N: \; {n \choose k} 2^{1-{k\choose 2}} < 1 \}$. Note that $$\frac{{n+1 \choose k}}{{n \choose k}} = \frac{n+1}{n+1-k}$$ so it suffices to prove that $N(k)/k \to \infty$ as $k \to \infty$. But for any finite $c > 1$, $$\log {c k \choose k} \sim (c \log c - (c-1) \log (c-1)) k = o(k^2)$$ so $${c k \choose k} 2^{1 - {k \choose 2}} \to 0 \ \text{as}\ k \to \infty$$ and thus for sufficiently large $k$ we must have $N(k) > c k$.
• Okay, why does $N(k)/k\to\infty$ suffice? Jun 29, 2018 at 18:03
• 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 20:41
• 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 20:58