Answer
The sum is $3$. See the graph below:
Work Step by Step
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.
