$a_n=\left(\frac{1}{2}\right)^{n-1} \ \ \text{Sum}=\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2$

$a_n=16\left(\frac{1}{4}\right)^{n-1} \ \ \text{Sum}=\frac{a_1}{1-r}$

$\text{Sum}=\frac{a}{1-r}$

The sum would be ∞ because r > 1.