Problem proving inequality $\frac{4^n}{n+1}\leq \frac{(2n)!}{(n!)^2}$

Question

I skip the base case $n=0$ because it's obvious.

I know that this is very equivalent to: Prove by induction: $\frac{2^{2n}}{n+1}<\frac{(2n)!}{(n!)^2},n>1$

But I try to learn some tricks so maybe you can help me?

Assumption: $\frac{4^n}{n+1}\leq \frac{(2n)!}{(n!)^2}\iff 4^n\leq\frac{(n+1)(2n)!}{(n!)^2}$ for some n.

Step: $n\to n+1$

$4^{n+1}=\frac{(n+2)(2n+2)!}{((n+1)!)^2}$

I begin with manipulating the LHS:

$4^{n+1}=4\cdot4^n\leq ???$

$???\leq\frac{(2n)!(2n+2)(2n+1)(n+1)}{(n!(n+1))}$

$=\frac{(n+1)(2n+2)!}{((n+1)!)^2}$

Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.

Can you help me?

Answer 1

$$\frac{4^{n+1}}{(n+2)} < \frac{(2n)!}{(n!)^2}\frac{4(n+1)}{(n+2)}$$ Now $$\frac{(2n)!}{(n!)^2}\frac{4(n+1)}{(n+2)} = \frac{(2n+2)!}{((n+1)!)^2}\frac{4(n+1)^3}{(n+2)(2n+1)(2n+2)}$$ So $$\frac{4(n+1)^3}{(n+2)(2n+1)(2n+2)} <1 $$ implies $$\frac{(2n+2)!}{((n+1)!)^2}\frac{4(n+1)^3}{(n+2)(2n+1)(2n+2)} <\frac{(2n+2)!}{((n+1)!)^2} $$, for $n \gt 1$.