calculus - Sum of the series:$frac{1}{1cdot 2}+frac{1cdot3}{1cdot2cdot3cdot4}+cdots$

I have a question where i couldn't find any clue.
The question is

$$\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+\cdots$$

I could get the general term as $t_n=\frac{1\cdot3\cdot5\cdot7\cdots(2n-1)}{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdots2n}$

I have also tried it to form the sequence in the telescopic form.But couldn't get. Any hint will be appreciated.

Answer

$$t_n=\frac{1\cdot3\cdot5\cdot7\dots(2n-1)}{1\cdot2\cdot3\cdot4\cdot5\cdot6\dots2n}=\frac{(2n-1)!!}{(2n-1)!!(2n)!!}=\frac{1}{(2n)!!}=\frac{1}{2^n n!}$$

Therefore this sum does not go to infinity.

Invoking $e^x = \sum^{\infty}_{n=0} \frac{x^n}{n!}$, we can compute the sum: $$\sum^{\infty}_{n=1} \frac{(\frac{1}{2})^n}{n!} = \sum^{\infty}_{n=0} \frac{(\frac{1}{2})^n}{n!}-1 = e^{\frac{1}{2}}-1$$