Answer
$\{a_{n}\}$ is bounded and monotonic$\Rightarrow$ it is convergent.
Work Step by Step
$\{a_{n}\}$ is bounded below by 0, as all terms are positive.
$\{a_{n}\}$ is bounded above by 1, because in
$\displaystyle \frac{2^{n}-1}{3^{n}},$
the numerator is smaller than the denominator (and both are positive).
If we can show that $\{a_{n}\}$ is monotonic, we can deduce by Th.6 that $\{a_{n}\}$ is convergent.
\begin{align*}
a_{n+1}-a_{n}&=\displaystyle \frac{2^{n+1}-1}{3^{n+1}}-\frac{2^{n}-1}{3^{n}} \\
& =\displaystyle \frac{2^{n+1}-1}{3^{n+1}}-\frac{3(2^{n}-1)}{3(3^{n})} \\
& =\displaystyle \frac{2^{n+1}-1-(3\cdot 2^{n+1}-3)}{3^{n+1}}\\
& =\displaystyle \frac{2-2^{n+1}}{3^{n+1}}\leq 0\\
a_{n+1}&\leq a_{n} \\ \end{align*}
It is monotonic. Thus, it is convergent.