Answer
Does not make sense.
Work Step by Step
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.
