Ramsey lower bounds

Question

I'm doing some self study on Ramsey graph theory. One of the first theorems concerning lower bounds shows that if $${n \choose k} \cdot \frac {1}{2^{( \frac {k}{2}-1)}} <1$$ then $N(k,k)> n$

In the derivation of a later result, the claim is made that due to stirlings formula, $${n \choose k} \cdot \frac {1}{2^{( \frac {k}{2}-1)}} <1$$

is implied by $$n^k \leq \left ( \frac {k}{e} \right )^k2^{\frac {(k^2-k)}{2}+1}, \; \; \text {or} \; \; n \leq \frac{k}{e \sqrt 2}2^{\frac {k}{2}}$$

The author claims both are obvious, any insight would be appreciated.

Answer 1

In the lower bound you quote it should be: $2^{\binom{k}{2}-1}$.
Sterling's formula implies the standard bound: $\binom{n}{k}\leq\left(\frac{en}{k}\right)^k $, which gives your result.

Answer 2

I believe that there are some mistakes there. The first equation should be $${n\choose k}\!\cdot \frac1{2^{\color{red} {{k\choose 2}}-1}}<1.$$ In the next equations, < should be replaced with >.(that's why you can use them as lower bounds.)

Now for the first inequality :$$ \begin{align} {n\choose k}=& \frac {n\cdot (n-1)...\cdot (n-(k-1))} {k!}\\&\text{so} \;n^k>{n\choose k}k!\end{align} $$And as the first equation governs,$${n\choose k}\! >2^{{k \choose 2}-1}. \\ 2^{{k \choose 2}-1}=2^{\frac{k(k-1)}2-1}=2^{\frac{k^2-k}2-1}. \\k!>\left(\frac ke\right)^k \sqrt{2\pi k}\;>\left(\frac ke\right)^k.\\\text{hence} \;\;n^k>\left(\frac ke\right) ^k2^{\frac{k^2-k}2-1}.$$Now to the second inequality. $$n=\left(n^k\right) ^{1/k}>\left(2^{\frac{k^2-k}2-1}k!\right)^{1/k}=2^{(\frac {k^2-k}2-1)\frac1k}\cdot k! ^{1/k}.\\2^{(\frac{k^2-k}2-1)\frac1k}=2^{\frac{k^2-k}{2k}-\frac1k}=2^{\frac k2-\frac12-\frac1k}\\(k!)^{1/k} >\left(\left(\frac ke\right)^k\sqrt{2\pi k}\right)^{1/k}=\frac ke\left(\sqrt{\pi2k}\right)^{1/k}=\frac ke(\pi 2k)^{\frac1{2k}}=\frac ke\cdot 2^{\frac1{2k}}(\pi k)^{\frac1{2k}}\\\text{combine}\;2^{\frac1{2k}}\;\text{with above}\;2^{\frac k2-\frac12-\frac1k}=2^{\frac k2-\frac12-\frac1k+\frac1{2k}}=2^{\frac k2-\frac12-\frac1{2k}}=\frac{2^{k/2}}{2^{1/2}2^{1/(2k)}}$$now everything together:$$n>\frac{k2^{k/2}\cdot(\pi k)^{1/(2k)}}{e\sqrt 2\cdot 2^{1/(2k)}}>\frac{k2^{k/2}}{e\sqrt2}$$The claim that all that is obvious, is obviously false.