prove the divergence How to show that the following series is divergent?
$$\sum_{n=1}^\infty \frac{4^n (n!)^2}{(2n)!}.$$
Stirling's approximation easily implies that the series is equivalent to $\sum \sqrt{n},$ hence diverges.
Can we solve it without using this result?
Thanks
 A: Note that the central binomial coefficient $\binom{2n}{n}$ is the largest one in the expansion of $(1+1)^{2n}$. Hence
$$\frac{(2n)!}{(n!)^2}=\binom{2n}{n}\leq  (1+1)^{2n}=4^n\implies \frac{4^n(n!)^2}{(2n)!}\geq 1.$$
A: Yes. Recall that $$
 \frac{(2n)!}{n!^2} = \binom{2n}{n}
$$
from which 
$$
\frac{(2n)!}{n!^2} \cdot \frac{1}{4^n} = \binom{2n}{n} \frac{1}{2^{2n}} = \mathbb{P}\{ X = n\} \leq 1
$$
where $X \sim\operatorname{Bin}(2n,1/2)$. This implies
$$
\frac{1}{\mathbb{P}\{ X = n\}} \geq 1
$$
leading to the divergence.
A: By Raabe's test
$$n\left(\frac{a_n}{a_{n+1}}-1\right)=n\left(\frac{4^n (n!)^2}{(2n)!}\frac{(2n+2)!}{4^{n+1} ((n+1)!)^2}-1\right)=n\left(\frac{(2n+2)(2n+1)}{4(n+1)^2}-1\right)=n\frac{4n^2+6n+2-4n^2-8n-4}{4n^2+8n+4}=\frac{-2n^2-2n}{4n^2+8n+4}\to -\frac12<1$$
thus the series diverges.
A: Hint: $(a+b)! \leq 2^{a+b} a!b!$, so $(2n)! = (n+n)! \leq 2^{n+n}n!n! = 4^n (n!)^2$.
The first inequality follows from the binomial theorem:
$$ 2^{a+b} = (1+1)^{a+b} = \sum_{k=0}^{a+b} \binom{a+b}{k} = \sum_{k=0}^{a+b} \frac{(a+b)!}{(a+b-k)!k!} \geq \frac{(a+b)!}{a!b!}, $$
where the bound in the final step follows by choosing $k=b$.
