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.