Let the first term be \(a\) and the common ratio be \(r\).

The third term is \(ar^2 = \frac{1}{3}\) and the fourth term is \(ar^3 = \frac{2}{9}\).

Divide the fourth term by the third term to find \(r\):

\(\frac{ar^3}{ar^2} = \frac{2/9}{1/3}\)

\(r = \frac{2}{3}\)

Substitute \(r = \frac{2}{3}\) into \(ar^2 = \frac{1}{3}\):

\(a \left( \frac{2}{3} \right)^2 = \frac{1}{3}\)

\(a \cdot \frac{4}{9} = \frac{1}{3}\)

\(a = \frac{3}{4}\)

The sum to infinity \(S_\infty\) is given by:

\(S_\infty = \frac{a}{1 - r}\)

\(S_\infty = \frac{\frac{3}{4}}{1 - \frac{2}{3}}\)

\(S_\infty = \frac{\frac{3}{4}}{\frac{1}{3}}\)

\(S_\infty = \frac{3}{4} \times 3 = 2 \frac{1}{4}\)