Summation involving double factorials How can we evaluate exact infinite sum of the following?
$${ \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}}$$
where 
$k!! = \begin{cases} 2\cdot4\cdot6 . . . k, & \text{if $k$ is even} \\ 1\cdot3\cdot5 . . . k, & \text{if $k$ is odd} \end{cases}$
What approach do we need to solve such type of summations?
My attempt:
\begin{align}
\sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}
 &= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!}
{(2r+3)!!(2r+2)!!} \\
 &= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cdot 2^{r+1}\cdot (r+1)!}{(2r+3)!} \\
 &= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!}
\end{align}
I also tried using the generalized binomial theorem but it doesn't seem to follow the pattern.
EDIT: I suspect it might be solved using telescopic method, but can't figure out how.

Starting with 

My attempt:
  \begin{align}
S = \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}
 &= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!}
{(2r+3)!!(2r+2)!!} \\
 &= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cdot 2^{r+1}\cdot (r+1)!}{(2r+3)!} \\
 &= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!}
\end{align}

One can continue as follows:
\begin{align}
S &= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!} \\
&= 2 \, \sum_{r=1}^{\infty} \frac{\Gamma(r+1) \, \Gamma(r+2)}{ \Gamma(2r + 4) } \, 4^{r} \\
&= \frac{2}{3!} \, \sum_{r=1}^{\infty} \frac{(1)_{r} \, (2)_{r}}{4^{r} \, (2)_{r} \, \left(\frac{5}{2}\right)_{r} } \, 4^{r} \\ 
&= \frac{1}{3} \, \sum_{r=1}^{\infty} \frac{(1)_{r} \, (1)_{r}}{r! \, \left(\frac{5}{2}\right)_{r} } = 2 \left[ {}_{2}F_{1}\left(1, 1; \frac{5}{2}; 1\right) -1 \right] \\
&= \frac{1}{3} \left[\frac{\Gamma(5/2) \, \Gamma(1/2)}{\Gamma^{2}(3/2)} - 1 \right] = \frac{2}{3}.
\end{align}
