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xFioraMstr18
xFioraMstr18
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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..

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..

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.

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xFioraMstr18
xFioraMstr18
  • 276
  • 1
  • 10

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

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xFioraMstr18
xFioraMstr18
  • 276
  • 1
  • 10

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

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..