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A well known proof by Erdős shows a lower bound on the Ramsey number $r(k,k)$ using the probabilistic method. The theorem goes thusly:

Let $n,\, k\in\mathbb{N}$ such that ${n \choose k}2^{1-{k \choose 2}}<1$. Then $r\left(k,k\right)>n$.

In particular, if $k\geq3$ then $r\left(k,k\right)>\lfloor2^{\frac{k}{2}}\rfloor$

I understand the proof of the first part, but I'm having trouble with the algebra in the case where $k\geq3$. It goes like this.

${n \choose k}2^{1-{k \choose 2}}<\frac{n^{k}}{k!}2^{1-\frac{k}{2}-\frac{k^{2}}{2}}=\frac{2^{1+\frac{k}{2}}}{k!}\left(\frac{n}{2^{\frac{k}{2}}}\right)^{k}\leq\frac{2^{1+\frac{k}{2}}}{k!}<1$

Could someone give me a hint how best to read this part?

Related question: Ramsey lower bounds

Related wiki entry: here

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π314
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  • Hmm.. is the fact that no one is answering an indication that there is something wrong with the question? – π314 Sep 13 '14 at 00:13
  • I just had a quick look at it, it is using the fact that in general ${n \choose k}\leq\frac{n^{k}}{k!}$. But the second term, should be in my opinion ${n \choose k}2^{1-{k \choose 2}}<\frac{n^{k}}{k!}2^{1+\frac{k}{2}-\frac{k^{2}}{2}}$ and not ${n \choose k}2^{1-{k \choose 2}}<\frac{n^{k}}{k!}2^{1-\frac{k}{2}-\frac{k^{2}}{2}}$ – Marco Bellocchi Feb 20 '18 at 23:02

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