Problem 648

Find the sum of the following infinite geometric series, if it exists.

$\frac{1}{2}+\left ( \frac{-1}{4} \right )+\frac{1}{8}+\left ( \frac{-1}{16} \right )+...$

Solution:-

S = $\frac{a}{1-r}$ , where r = common ratio (r ≤ 1) and a = first term.

a = $\frac{1}{2}$ , r = -1/8

S = $\frac{\frac{1}{2}}{\left ( 1-\left ( \frac{-1}{8} \right ) \right )}$

S = $\frac{4}{9}$

So the sum of the infinite geometric series 4/9