Prove that for $n\ge 2$, $(n + 1)^n n! < (2n)! < 4^n (n!)^2 .$

Question

Prove that for all real $n\ge 2$, the following inequalities hold: $$(n + 1)^n n! < (2n)! < 4^n (n!)^2 .$$

I have posted my proof below, if anyone has a shorter, or more interesting proof, please share :-)

Answer 1

First note we can re-write the left hand side as: $$(n+1)^n n!=1\cdot2\cdot3\cdots n\cdot \underbrace{(n+1)\cdot(n+1)\cdots(n+1)}_{n\text{ times}},$$ and the middle term as $$(2n)!=1\cdot2\cdot3\cdots n\cdot(n+1)\cdot (n+2)\cdots (n+(n-1))\cdot 2n.$$ The first $n+1$ terms are equal, and the remaining terms can be compared as follows: $$\begin{align} n+2 &> n+1,\\ n+3 &> n+1,\\ &\vdots\\ 2n-1&> n+1,\\ 2n &> n+1. \end{align}$$ It follows that $(2n)!>(n+1)^n n!$.

It remains to prove that $4^n(n!)^2>(2n)!$, we proceed by induction. For the base case it is clear that $4^2(2!)^2=64>24=(2\cdot 2)!$. Assume the statement is true for $n=k$, and consider $n=k+1$. The middle term becomes $$(2(k+1))!=(2k)! \cdot (2k+1) (2k+2),$$ and the right hand side is $$4^{k+1} ((k+1)!)^2=4^k(k!)^2\cdot 4(k+1)(k+1).$$ From the inductive hypothesis, we know that $4^k(k!)^2>(2k)!$, and clearly $$4(k+1)(k+1)=2(k+1)\cdot 2(k+1)=(2k+2)(2k+2)>(2k+1)(2k+2),$$ and the result follows.