Answer
Prove $\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}$ is $\Theta(1)$.
By theorem 5.2.4: $\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}} = \frac{1}{5} (1 + \frac{4}{5} + \frac{4^2}{5^2} + \cdots + \frac{4^n}{5^{n}}) = \frac{1}{5}(\frac{\frac{4}{5}^{n+1}-1}{\frac{4}{5}-1}) = -(\frac{4}{5}^{n+1}-1)=1-\frac{4}{5}^{n+1}$.
For $n>0$,
$1-\frac{4}{5}^{n+1} \leq 1$.
By the transitive property:
$\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}} \leq 1$.
For $n>0$,
$\frac{1}{5}(1) \leq \frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}$.
Since all terms are positive:
$\frac{1}{5}|(1)| \leq |\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}| \leq |1|$.
Thus for A=1/5, B=1, k=1,
$A|(1)| \leq |\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}| \leq B|1|$ for all $x>k$.
Thus $\frac{1}{5} + \frac{4}{5^2} + \frac{4^2}{5^3} + \cdots + \frac{4^n}{5^{n+1}}$ is $\Theta(1)$.
Work Step by Step
Recall the definition of $\Theta$-notation:
$f(x)$ is $\Theta(g(x))$ iff there exist positive real numbers A, B, k, such that $A|g(x)| \leq |f(x)| \leq B|g(x)|$ for all $x>k$.
Recall theorem 5.2.3: Sum of a geometric sequence:
For any real number $r$ except 1, and any integer $n \geq 0$,
$\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$.
In this case $r=\frac{4}{5}$.