## Precalculus (6th Edition) Blitzer

We can observe that the given series is an infinite geometric series with $a=3$ and $r=\frac{1}{3}$. Thus \begin{align} & {{S}_{n}}=\frac{a}{\left( 1-r \right)} \\ & =\frac{3}{\left( 1-\frac{1}{3} \right)} \\ & =\frac{3}{\left( \frac{2}{3} \right)} \\ & =\frac{9}{2} \end{align} Therefore, we cannot find the sum of the infinite series by addition.