<s>[INST] 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?
 [/INST] You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)} &=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2(2n+2)}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2 \cdot (2n+2)}\\
& =\frac{(2n)!}{2^{2n} (n!)^2 \cdot (2n+2)}\\
& \sim \frac{1}{\sqrt{\pi n}\cdot (2n+2)}\\
& \sim \frac{1}{2\sqrt{\pi}\cdot n^{\Large\frac 32}}, \quad \text{for} \, n \, \text{great}
\end{align}
$$
where we have use Stirling's approximation, then you easily conclude to the convergence of the series.
</s>