TITLE: Sum of the series $\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\dots$
QUESTION [7 upvotes]: 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  .

REPLY [15 votes]: While investigating the question, I found an easy answer., via telescoping series.:
rearrange the fractions as 
$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}=\frac{1\cdot(4-3)}{2\cdot4}+\frac{1\cdot3\cdot(6-5)}{2\cdot4\cdot6}+\cdots$$
$$\frac{1\cdot4}{2\cdot4}-\frac{1\cdot3}{2\cdot4}+\frac{1\cdot3\cdot6}{2\cdot4\cdot6}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\cdots=\frac{1}{2}$$