prove the divergence

Question

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

Answer 1

Note that the central binomial coefficient $\binom{2n}{n}$ is the largest one in the expansion of $(1+1)^{2n}$. Hence $$\frac{(2n)!}{(n!)^2}=\binom{2n}{n}\leq (1+1)^{2n}=4^n\implies \frac{4^n(n!)^2}{(2n)!}\geq 1.$$

Answer 2

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 $k=b$.

Answer 3

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.

Answer 4

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.