+70 Evaluate the infinite geometric series: \(\frac13+\frac16+\frac{1}{12}+\frac{1}{24} \dots\)
+9673 First term is 1/3.
Common ratio is (1/6)/(1/3) = 1/2.
Using the formula sum = a/(1 - r), sum = \(\dfrac{\frac13}{1 - \frac12} = \dfrac23\)
+150 Observe that we can say
\(\displaystyle\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \cdots = \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{4}+\cdots\right) = \frac{1}{3}\sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n.\)
Given this, our common ratio \(r\) is \(r = \frac{1}{2}\). Then, by the well-known infinite geometric series formula, we have
\(\displaystyle \frac{1}{3} \sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n = \frac{1}{3} \cdot\frac{1}{1-\frac{1}{2}} = \frac{2}{3}.\)
