Answer
$\{a_{n}\}$ is convergent.
Work Step by Step
$ a_{n}=4+(\displaystyle \frac{3}{4})^{n}\geq 4\quad$ ... bounded below by $4$.
$(\displaystyle \frac{3}{4})^{n}$is the greatest when n is the smallest (n=1)
$\{a_{n}\}$ is bounded above by $4+\displaystyle \frac{3}{4}$
$\{a_{n}\}$ is bounded. If we can show that it is monotonic, we can apply Th.6.
\begin{align*}
a_{n+1}-a_{n}&=4+(\displaystyle \frac{3}{4})^{n+1}-(4+(\frac{3}{4})^{n}) \\
& =(\displaystyle \frac{3}{4})^{n+1}-(\frac{3}{4})^{n}\\
& =(\displaystyle \frac{3}{4})^{n}(\frac{3}{4}-1)\leq 0\\
a_{n+1}&\leq a_{n} \\ \end{align*}
$\{a_{n}\}$ is nonincreasing, that is, monotonic. Th.6 applies.
$\{a_{n}\}$ is convergent.
