Answer
Monotonic (nondecreasing) and bounded from above.
Work Step by Step
A sequence is monotonic if it is nondecreasing or nonincreasing.
\begin{align*}
a_{n+1}-a_{n}&=(2-\displaystyle \frac{2}{n}-\frac{1}{2^{n}})-(2-\frac{2}{n+1}-\frac{1}{2^{n+1}}) \\
& =(\displaystyle \frac{2}{n+1}-\frac{2}{n})+(\frac{1}{2^{n+1}}-\frac{1}{2^{n}}) \\
& =\displaystyle \frac{2n-2(n+1)}{n+1}+\frac{1-2}{2^{n+1}} \\
& =\displaystyle \frac{-1}{n+1}+\frac{-1}{2^{n+1}} \\
& =-(\displaystyle \frac{1}{n+1}+\frac{1}{2^{n+1}})\leq 0 \\
a_{n+1}& \leq a_{n} \end{align*}
Thus, $a_{n}$ is monotonic (nondecreasing).
Also,
$a_{n}=2-(\displaystyle \frac{2}{n}+\frac{1}{2^{n}})\lt 2$, so $\{a_{n}\}$ is bounded from above.
