Prove by induction: $\frac{2^{2n}}{n+1}<\frac{(2n)!}{(n!)^2},n>1$

Question

Prove by induction: $\frac{2^{2n}}{n+1}<\frac{(2n)!}{(n!)^2},n>1$

For $n=2$ equality holds.

For $n=k$

$$\frac{2^{2k}}{k+1}<\frac{(2k)!}{(k!)^2}$$

For $n=k+1$

$$\frac{2^{2k+2}}{k+2}<\frac{(2k+2)!}{((k+1)!)^2}$$

How to prove this?

Answer 1

You need to show that: $$\frac{4(k+1)}{k+2}\leq \frac{(2k+2)(2k+1)}{(k+1)^2}$$

or, equivalently:

$$2(k+1)^2\leq (2k+1)(k+2)$$

Answer 2

$$\begin{align} \frac{2^{2n+2}}{k+2}&=\frac{2^{2n}}{k+1}\frac{4(k+1)}{k+2}\\\\ &<\frac{(2k)!}{(k!)^2}\frac{4(k+1)}{k+2}\\\\ &=\left(\frac{(2k+2)!}{((k+1)!)^2}\right)\left(\frac{4(k+1)(k+1)^2}{(k+2)(2k+2)(2k+1)}\right)\\\\ &=\left(\frac{(2k+2)!}{((k+1)!)^2}\right)\left(\frac{2(k+1)^2}{(k+2)(2k+1)}\right)\\\\ &=\left(\frac{(2k+2)!}{((k+1)!)^2}\right)\left(\frac{2k^2+4k+4}{2k^2+5k+2}\right)\\\\ &<\frac{(2k+2)!}{((k+1)!)^2} \end{align}$$

for $k\ge 2$, as was to be shown!