Convergent or divergent $\sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}}$ \begin{align}
  & \sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}} \\ 
 & \text{ordering} \\ 
 & a_{n}=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)\cdot 1}{2\cdot 4\cdot 6...(2n-2)(2n)(2n+2)}= \\ 
 & \underbrace{\left( 1-\frac{1}{2} \right)\left( 1-\frac{1}{4} \right)...\left( 1-\frac{1}{2n} \right)}_{n\text{ times}}\frac{1}{(2n+2)} \\ 
 & \text{clearly} \\ 
 & \left( 1-\frac{1}{2} \right)^{n}\frac{1}{(2n+2)}\le a_{n}\le \left( 1-\frac{1}{2n} \right)^{n}\frac{1}{(2n+2)} \\ 
 & \text{Root test} \\ 
 & \sqrt[n]{\left( 1-\frac{1}{2} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}}\le \sqrt[n]{a_{n}}\le \sqrt[n]{\left( 1-\frac{1}{2n} \right)^{n}}\frac{1}{\sqrt[n]{(2n+2)}} \\ 
 & n\to \infty  \\ 
 & \frac{1}{2}\le \underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{a_{n}}\le 1 \\ 
 & \text{nothing :(} \\ 
 & \text{Ratio test} \\ 
 & \frac{a_{n}}{a_{n-1}}=\frac{1\cdot 3\cdot 5...(2n-3)(2n-1)}{2\cdot 4\cdot 6...(2n)(2n+2)}\centerdot \frac{2\cdot 4\cdot 6...(2(n-1)+2)}{1\cdot 3\cdot 5...(2(n-1)-1)}=\frac{2n-1}{2n+2} \\ 
 & \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{a_{n-1}}=1 \\ 
 & \text{nothing again} \\ 
\end{align}

Any suggestions?

An explicit computation is even better. We have:
$$\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{(2n+2)\,4^n}\binom{2n}{n}=\frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1}$$
hence your series is telescopic and we have:
$$\sum_{n=1}^{N}\frac{(2n-1)!!}{(2n+2)!!}=\frac{1}{2}-\frac{1}{4^{N+1}}\binom{2N+2}{N+1}=\frac{1}{2}+O\left(\frac{1}{\sqrt{N}}\right).$$
In order to prove the last asymptotics, notice that:
$$\frac{1}{4^n}\binom{2N}{N}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos(x)^{2N}dx\leq\frac{2}{\pi}\int_{0}^{\pi/2}e^{-Nx^2}\,dx\leq\frac{1}{\sqrt{\pi N}}.$$
