Series question,related to telescopic series, 1/2*4+ 1*3/2*4*6+ 1*3*5/2*4*6*8 ...infinity The series is $$\frac{1}{2*4}+ \frac{1*3}{2*4*6}+ \frac{1*3*5}{2*4*6*8}+....$$ 
It continues to infinity.I tried multiplying  with $2$ and dividing each term by$(3-1)$,$(5-3)$ etc,starting from the second term which gives me $$\frac{1}{8} -\frac{1}{4*6} -\frac{1}{4*6*8}-\frac{1}{4*6*8*10}$$.
Also if anybody is wondering the answer 0.5

The $r$th term $\displaystyle\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$  where integer $r\ge1$
As the number of terms in the numerator $=$ the number of terms in the denominator $-1,$
If $\displaystyle S=\sum_{n=1}^\infty\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)},$
Multiplying the numerator by the previous term of $1,3,5,\cdots$ (Arithmetic Series)
$\displaystyle -S=\sum_{n=1}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$
$\displaystyle=-\frac{-1}2+\sum_{n=0}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$
Now, $\displaystyle\sum_{n=0}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}=(1-1)^{\frac12}-1$
Now follow Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
