Prove that $\frac{(2n)!}{(n!)^2} \gt \frac{4^n}{2n+1} \forall n \in \mathbb{N}$

Question

I have proved it using induction.

My try:

I rewrote the inequality as $ (2n+1)! \gt (2^n.n!)^2$

Proof by induction:

Base case:

for $n=1$ , we obtain $3! \gt 4$ which is true. $\implies$$P(1)$ is true.

Induction step:

Assuming that the statement is true for a natural number $k$, we obtain $ (2k+1)! \gt (2^k.k!)^2$.

Now, for $n=k+1$,

$ (2k+3)! \gt (2^{k+1}.(k+1)!)^2$

$\implies$$(2k+3).(2k+2).(2k+1)! \gt 4(k+1)^2 (2^k.k!)^2$

Since we know that $ (2k+1)! \gt (2^k.k!)^2$ is true,

if $(2k+3).(2k+2) \gt 4(k+1)^2$ then out claim is true.

$\implies$ $(2k+3).(2k+2) \gt (2k+2)(2k+2)$

$\implies$ $(2k+3) \gt (2k+2)$

which is trivially true.

Which means $P(k) \implies P(k+1)$. Hence proved.

But my question is, how do you prove this using binomial theorem (or any other method)?

Answer 1

By the binomial thereom, we have $$ 4^n = (1 + 1)^{2n} = \sum\limits_{k = 0}^{2n} {\binom{2n}{k}} < \sum\limits_{k = 0}^{2n} \binom{2n}{n} = (2n + 1)\binom{2n}{n} = (2n + 1)\frac{{(2n)!}}{{(n!)^2 }}, $$ since $\binom{2n}{n}$ is maximal among all the binomial coefficients $\binom{2n}{k}$.

Answer 2

When you face problems involving factorials, think about Stirling approximation.

Considering $$A=\frac{(2n)!}{(n!)^2} \implies \log(A)= \log ((2 n)!)-2 \log (n!)$$ Use Stirling approximation twice to get $$\log(A)=n \log (4)-\frac{1}{2} \log \left({\pi n}\right)+O\left(\frac{1}{n}\right)$$ Now, use $$A=e^{\log(A)}=\frac{4^n }{\sqrt{n\pi }}$$ It just remains to compare $\sqrt{n\pi }$ and $(2n+1)$.