## Discrete Mathematics with Applications 4th Edition

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 $\Theta(1)$.
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 $r=\frac{4}{5}$.