## Precalculus (6th Edition) Blitzer

The sum is $3$. See the graph below:
Consider the given function; $f\left( x \right)=\frac{2\left[ 1-{{\left( \frac{1}{3} \right)}^{x}} \right]}{1-\frac{1}{3}}$ The infinite geometric series is: $2+2\left( \frac{1}{3} \right)+2{{\left( \frac{1}{3} \right)}^{2}}+2{{\left( \frac{1}{3} \right)}^{3}}+\cdots$ From the graph, the horizontal asymptote of the given function is $y=3$. Consider the given series, $2+2\left( \frac{1}{3} \right)+2{{\left( \frac{1}{3} \right)}^{2}}+2{{\left( \frac{1}{3} \right)}^{3}}+\cdots$ The first term of the series is ${{a}_{1}}=2$. And the common ratio is \begin{align} & r=\frac{1}{3} \\ & <1 \end{align} Use the formula for the sum of an infinite geometric series with the first term ${{a}_{1}}$ and common ratio r, $S=\frac{{{a}_{1}}}{1-r}$ Substituting ${{a}_{1}}=2$ and $r=\frac{1}{3}$, the sum of the given infinite geometric series is \begin{align} & S=\frac{2}{1-\left( \frac{1}{3} \right)} \\ & =\frac{2}{\frac{2}{3}} \\ & =3 \end{align} So, the sum of the given infinite series is 3.