Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 11 - Analysis of Algorithm Efficiency - Exercise Set 11.4 - Page 764: 40


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 $n \geq 0$, $\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$.
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