Question

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

Answer 1

we can verify for n = 2 as in part (i)
Assume p(n) i. e.
$\frac{\left(2n\right)!}{(n!{)}^2}>\frac{4n}{2n+1}$
p(n + 1) = $\frac{(2n+2)!}{\left[\right(n+1)!{]}^2}$
=$\frac{(2n+2)(2n+1)}{(n+1{)}^2}\frac{\left(2n\right)!}{(n!{)}^2}$
$=\frac{2(2n+1)}{n+1}p\left(n\right)>\frac{2(2n+1)}{n+1}\cdot \frac{4n}{2n+1}$
$=\frac{8n}{n+1}$
We have prove p( n +1) > $\frac{4(n+1)}{2n+3}$
or $\frac{8n}{n+1}>\frac{4(n+1)}{2n+1}$
or $2n(2n+3)-$ $(n^2$ $+2n+1)>0$
or $3n^2$ $+3n+n-1>0$
or 3$n(n+1)+(n-1)>0$
Above is true for n > 1