Answer
Prove $\frac{2n}{3} + \frac{2n}{3^2} + \frac{2n}{3^3} + \cdots + \frac{2n}{3^n}$ is $\Theta(n)$.
By theorem 5.2.3: $\frac{2n}{3} + \frac{2n}{3^2} + \frac{2n}{3^3} + \cdots + \frac{2n}{3^n} = \frac{2n}{3}(1+ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots + \frac{1}{3^{n-1}}) = \frac{2n}{3}(\frac{(\frac{1}{3})^n-1}{\frac{1}{3}-1}) = n(1-(\frac{1}{3})^n)$.
For $n>1$,
$\frac{2}{3} \leq 1- (\frac{1}{3})^n \leq 1$
so $n(\frac{2}{3}) \leq n(1- (\frac{1}{3})^n) \leq n$.
By the transitive property:
$n(\frac{2}{3}) \leq \frac{2n}{3} + \frac{2n}{3^2} + \frac{2n}{3^3} + \cdots + \frac{2n}{3^n} \leq n$.
Since all terms are positive:
$(\frac{2}{3})|n| \leq |\frac{2n}{3} + \frac{2n}{3^2} + \frac{2n}{3^3} + \cdots + \frac{2n}{3^n}| \leq 1|n|$.
Thus for A=2/3, B=1, k=1,
$A|n| \leq |\frac{2n}{3} + \frac{2n}{3^2} + \frac{2n}{3^3} + \cdots + \frac{2n}{3^n}| \leq B|n|$ for all $n>k$.
Thus $\frac{2n}{3} + \frac{2n}{3^2} + \frac{2n}{3^3} + \cdots + \frac{2n}{3^n}$ is $\Theta(n)$.
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 $n \geq 0$,
$\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$.
