Answer
$$\frac{125}{18}$$
Work Step by Step
Given $$\frac{25}{9}+\frac{5}{3}+1+\frac{3}{5}+\frac{9}{25}+\frac{27}{125}+\cdots=\sum_{n=0}^{n=\infty} \frac{25}{9}\left(\frac{3}{5}\right)^{n} $$
Since the series is a geometric series with $|r|= \frac{3}{5}<1$, then the series converges and has the sum
\begin{align*}
S &=\frac{a_1}{1-r} \\
&=\frac{\frac{25}{9}}{1-\frac{3}{5}} \\
&=\frac{125}{18}
\end{align*}
