Convergence of the series $\sum_{n=1}^\infty \frac{(2n)!!}{(2n+1)!!} $ 
Study the convergence of the next series: $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+1)!!} $$

My solution: since $$\frac{(2n)!!}{(2n+2)!!} \leq \frac{(2n)!!}{(2n+1)!!}$$
forall $n \in \mathbb{N}$ and since $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+2)!!} = \sum_{n=1}^\infty \frac{1}{2n+2} = \infty$$
then the first series diverges. Is there anything wrong? Thanks in advance.

Yes. What you have done is correct. You can attempt to quantify the divergence a bit better. Note that $(2n)!! = 2^n n!$ and $(2n+1)!! = \dfrac{(2n+2)!}{2^{n+1} (n+1)!}$. Hence,
$$\dfrac{(2n)!!}{(2n+1)!!} = 2^{2n+1} \dfrac{(n+1)! n!}{(2n+2)!} = \dfrac{2^{2n+1}}{\dbinom{2n+2}{n+1} \cdot (n+1)}$$
We have that $\dbinom{2n+2}{n+1} \sim \dfrac{4^{n+1}}{ \sqrt{\pi(n+1)}}$. Hence,
$$\dfrac{2^{2n+1}}{\dbinom{2n+2}{n+1} \cdot (n+1)} \sim \dfrac{2^{2n+1} \sqrt{\pi}}{4^{n+1} \sqrt{n+1}} = \dfrac{\sqrt{\pi}}2 \cdot \dfrac1{\sqrt{n+1}}$$
Hence, $$\sum_{n=1}^{\infty}\dfrac{(2n)!!}{(2n+1)!!} \,\,\,\,\,\,\,\,\text{diverges as} \,\,\,\,\,\, \dfrac{\sqrt{\pi}}2 \cdot \sum_{n=1}^{\infty}\dfrac1{\sqrt{n+1}}$$
