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

Question

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.

Answer 1

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 $N(k) > c k$.