JAMB 2004 · UME · Q20

Find the sum to infinity of the series $\frac12, \frac16, \frac1{18}, \dots$

Worked solution (try it first)

This is a G.P. with $a = \frac12$ and $r = \frac16 \div \frac12 = \frac13$ .
Since $r$ is between $-1$ and 1, $S_\infty = \dfrac{a}{1 - r}$ , and $1 - \frac13 = \frac23$ .
Dividing by $\frac23$ means multiplying by $\frac32$ : $S_\infty = \frac12 \times \frac32$
$= \frac34$ , option B.