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

Question

Using mathematical induction prove that:

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

I tried to use $n + 1$:

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

But still don't see anything to do after.

Have you noted that $$ \frac {(2(n + 1))!} {((n + 1)!)^2} = \frac{(2n+1)(2n+2)}{(n+1)^2}\cdot \frac{(2n)!}{(n!)^2} $$and tried to use the induction hypothesis on that expression? — Arthur, Nov 19 '15 at 19:37

score 0 · Answer 1 · answered Nov 19 '15 at 19:52

Actually you can do the following: in the induction step before you go to (n+1) notice that 4*(2n)!/(n!)^2 > 4*4^n/(n+1) > 4*4^n/(n+2)

your are deviding by (n+2) so you get a smaller number compared to a division by (n+1)

so you have: 4*(2n)!/(n!)^2 > 4^(n+1)/(n+2)

you can now easily show that for n sufficiently big you get.

(2(n+1))!/(n+1)!^2 > 4*(2n)!/(n!)^2

Good Luck

1 Answers1