Prove that $(1007!)^2 > \binom{2015}{1007}^2$

Question

Prove that $(1007!)^2 > \binom{2015}{1007}^2$.

This is equivalent to proving that $1007! > \binom{2015}{1007}$. We then get this is equivalent to $$(1007!)^2 > \dfrac{2015!}{1008!} = 2015 \cdot 2014 \cdots 1009.$$ How do we show this?

Answer 1

I will follow the comment of Maman (which I thank warmly), the first one that had the idea to show a more general statement: $n!>\binom{2n+1}{n}$.

We show by induction (what else?) that $n!>\binom{2n+1}{n}$ for $n\geq 8$ (8 comes from this evaluation wolframalpha).

For $n=8$ it is ok. Inductive step. For $n\geq 8$, $$(n+1)!=(n+1)n!>(n+1)\binom{2n+1}{n}\stackrel{?}{\geq}\binom{2n+3}{n+1}.$$ So we have to verify that $$\frac{(2n+1)!}{n! n!}\geq \frac{(2n+3)!}{(n+1)! (n+2)!},$$ that is $$(n+1)^2(n+2)\geq (2n+3)(2n+2)$$ which holds for $n\geq 8$.

Answer 2

You get $(1007!)^2$ compared to $\frac{2015!}{1008!}$. The second expression is a product of $1007$ integers. The first expression is the product of $2014$ integers. If you take them in carefully chosen pairs, you can have the product of each pair greater than each factor in the second expression.

Answer 3

Consider the ratios $$A_n=\frac{(2n+1)!}{n!^2(n+1)!}$$ The goal is to show that $$A_{1007}<1$$ To do so, compute $$A_8=\frac{2431}{4032}<1$$ and note that $$\frac{A_{n+1}}{A_n}=\frac{2(2n+3)}{(n+1)(n+2)}<1$$ for every $n\geqslant3$ since the difference of the denominator and the numerator is $$n^2-n-4\geqslant(n-3)(n+2)$$ This shows that $$A_n<1$$ for every $n\geqslant8$.