Answer
Monotonic and Bounded
Work Step by Step
$a_n = (\frac{2}{3})^n $
$a_{n+1} = (\frac{2}{3})^{n+1} = \frac{2}{3} \times (\frac{2}{3})^n=\frac{2}{3} \times a_n $
Therefore, $a_{n+1}\lt a_n$ for all n.
Or $a_1>a_2>...$
Hence the sequence is monotonic.
Also note that each term of the sequence is less than $a_1$, i.e $ \frac{2}{3}$. So it is bounded from above by $\frac{2}{3}$
Since none of the numbers is negative, and $\lim\limits_{n \to \infty}(\frac{2}{3})^n =0$, the sequence is bounded from below by 0
Hence, the sequence is monotonic and bounded.
