I'm reading Spencer's lectures on the probabilistic method. Using the Lovasz local lemma, we've shown that $R(k,k)>n$ if $$ 4{k \choose 2} {n \choose k-2} 2^{1-{k \choose 2}} < 1. $$
Now I'm trying to analyze the asymptotics: $$ 2k^2 \frac{(ne/k)^{k-2}}{\sqrt{2\pi(k-2)}}2^{1-k^2/2}<1 \\ n\sim \frac{k}{e} 2^{-(2-k^2/2)/(k-2)}=\frac{2}{e}k2^{k/2}. $$
However, Spencer (without explanation) says $n\sim \frac{\sqrt{2}}{e}k2^{k/2}$. Where did I go wrong?
The error comes from replacing $\binom{k}{2}$ with $k^2/2$. Leave it as $k(k-1)/2$.
Then the exponent on the 2, after dividing by $k-2$ becomes $${\frac{k(k-1)}{2}\cdot\frac{1}{k-2}}= (\frac{k}{2}-\frac{1}{2})\cdot(1 +\frac{2}{k} + O(1/k^2)) = \frac{k}{2} + \frac 12 + o(1).$$ If you do Erd\H{o}s's original argument and leave $\binom{k}{2}$ as $k(k-1)/2$ you will see where the $\sqrt{2}$ in the denominator comes from.