Sum of the series $\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\dots$ I am recently struck upon this question that asks to find the sum until infinite terms
$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+.....∞$$
I tried my best to get something telescoping or something useful, but I failed. I even made a recurrence as $t_n=t_{n-1}\frac{2n-1}{2n+2}$, but this question was expected to be done with simple logic of series (Also that the recurrence on solving gives a higher order charasteristic polynomial, which might be difficult to solve without calculator). So, thus I ended up being confused with this question. So, can anyone prode a small solution to this (might be easy) problem  .
 A: $$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}$$

Rewrite the sum as
\begin{align}
\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots
&=\sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!}\\
&=\sum^\infty_{n=0}\frac{(2n+1)!}{2^{2n+2}n!(n+2)!}\\
&=\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}\Bigg{|}_{x=1/2}
\end{align}
Differentiate once to get
\begin{align}
\frac{{\rm d}}{{\rm d}x}\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}
&=\sum^\infty_{n=0}\binom{2n+2}{n}x^{2n+1}\\
&=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{z}\sum^\infty_{n=0}\left[\frac{(x+xz)^{2}}{z}\right]^n{\rm d}z\\
&=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{z-x^2(1+z)^2}{\rm d}z\\
&=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{-x^2z^2+(1-2x^2)z-x^2}{\rm d}z\\
&=-\frac{1}{i2\pi x^2}\oint_{|z|=1}\frac{x(1+z)^2}{(z-r_+)(z-r_-)}{\rm d}z\\
&=-\frac{1}{x^2}{\rm Res}\left[\frac{x(1+z)^2}{(z-r_+)(z-r_-)},r_+\right]\\
&=-\frac{1}{x}\frac{(1+r_+)^2}{r_+ - r_-}
\end{align}
We know that $r_+=\dfrac{1-2x^2-\sqrt{1-4x^2}}{2x^2}$. Coupled with the fact that $r_+  - r_- =\dfrac{-\sqrt{1-4x^2}}{x^2}$, this gives us
\begin{align}
\sum^\infty_{n=0}\binom{2n+2}{n}x^{2n+1}
&=\frac{\sqrt{1-4x^2}}{4x^3}-\frac{1}{2x^3}+\frac{1}{4x^3\sqrt{1-4x^2}}
\end{align}
Now, integrate this expression once
\begin{align}
\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}
&=\frac{1-\sqrt{1-4x^2}}{4x^2}+C
\end{align}
It is easy to see that
\begin{align}
C
=\lim_{x\to 0}\frac{\sqrt{1-4x^2}-1}{4x^2}
=\lim_{u\to 1}\frac{u-1}{1-u^2}
=-\lim_{u\to 1}\frac{1}{1+u}
=-\frac{1}{2}
\end{align}
Finally, letting $x=\dfrac{1}{2}$ yields
\begin{align}
\sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!}
&=\frac{1-\sqrt{1-1}}{1}-\frac{1}{2}\\
&=\frac{1}{2}
\end{align}
A: While investigating the question, I found an easy answer., via telescoping series.:
rearrange the fractions as 
$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}=\frac{1\cdot(4-3)}{2\cdot4}+\frac{1\cdot3\cdot(6-5)}{2\cdot4\cdot6}+\cdots$$
$$\frac{1\cdot4}{2\cdot4}-\frac{1\cdot3}{2\cdot4}+\frac{1\cdot3\cdot6}{2\cdot4\cdot6}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\cdots=\frac{1}{2}$$
