Answer
$\{a_{n}\}$ is bounded and monotonic $\Rightarrow$ it is convergent.
Work Step by Step
$\{a_{n}\}$ is bounded above by 1, because in
$\displaystyle \frac{2^{n}-1}{2^{n}},$
the numerator is smaller than the denominator (and both are positive).
$\{a_{n}\}$ is bounded below by 0.
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}{2^{n+1}}-\frac{2^{n}-1}{2^{n}} \\
& =\displaystyle \frac{2^{n+1}-1}{2^{n+1}}-\frac{2(2^{n}-1)}{2(2^{n})} \\
& =\displaystyle \frac{2^{n+1}-1-(2^{n+1}-2)}{2^{n+1}}\\
& =\displaystyle \frac{1}{2^{n+1}}\geq 0\\
a_{n+1}&\geq a_{n} \\ \end{align*}
It is monotonic. Thus, it is convergent.
