Sum of the series:$\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\cdots$

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

Therefore this sum does not go to infinity.

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

The lowest form of each term is $\frac{(\frac{1}{2})^n}{n!}$. By invoking the Taylor series expansion of $e^x$ the answer is $e^{1/2}-1$.