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.
 A: Notice
$$\begin{align}
\frac{(2r)!!}{(2r+3)!!} 
&= \frac{2\cdot 4 \cdots \cdot 2r}{3\cdot 5 \cdots (2r+3)}= \frac13 \frac{\prod\limits_{s=0}^{r-1}(1+s)}{\prod\limits_{s=0}^{r-1}(\frac52+s)}\\
&= \frac13 \frac{\Gamma(r+1)/\Gamma(1)}{\Gamma(r+\frac52)/\Gamma(\frac52)}
= \frac12\frac{\Gamma(r+1)\Gamma(\frac32)}{\Gamma(r+\frac52)}\\
&= \frac12 \int_0^1 t^{(r+1)-1} (1-t)^{3/2-1} dt
= \frac12\int_0^1 t^r \sqrt{1-t}dt
\end{align}
$$
We have
$$\begin{align}\sum_{r=1}^\infty \frac{(2r)!!}{(2r+3)!!} &= 
\frac12\sum_{r=1}^\infty \int_0^1 t^r\sqrt{1-t} dt
= \frac12 \int_0^1 \left(\sum_{r=1}^\infty t^r\right)\sqrt{1-t} dt\\
&= \frac12 \int_0^1 \frac{t}{\sqrt{1-t}}dt = \frac23
\end{align}
$$
A: Let $A_r = \frac23 \frac45 \cdots \frac{2r}{2r+1}$. Then 
$$
\frac{(2r)!!}{(2r+3)!!} = A_r \cdot \frac1{2r+3}.
$$
Use $\frac1{2r+3} = 1 - \frac{2r+2}{2r+3}$. Then the sum telescopes. So, we are to evaluate 
$$
\sum_{r=1}^{\infty} (A_r - A_{r+1}).
$$
To complete the solution, we need the proof of $\lim_{r\rightarrow\infty} A_r = 0$. Your expressions and Stirling's formula will do the job. 
A: 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}
