Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.3 Infinite Series - 8.3 Exercises - Page 623: 59

Answer

$\dfrac{1}{9}$

Work Step by Step

Let us consider an infinite series $\Sigma_{n=k}^{n=\infty} a_n$. This series will converge to the sum $S$ when the sequence of its partial sums $\{S_n\}$ converge to $S$. That is, $S=\lim\limits_{N \to \infty} S_n$ and $S=\Sigma_{n=k}^{n=\infty} a_n$. This series will diverge when the limit does not exist and the series will diverge to infinity when the limit is infinite. Here, we have the sum $S$ is the limit of the sequence of partial sums which is equal to: $S=\lim\limits_{n \to \infty} S_n\\=\lim\limits_{n \to \infty} (\dfrac{1}{9} -\dfrac{1}{4n+1})\\=\lim\limits_{n \to \infty} \dfrac{1}{9}-\lim\limits_{n \to \infty} \dfrac{1}{4n+1}$ Because for all $n \gt 0$ , $\lim\limits_{x \to \infty} x^{-n}=0$ and $\lim\limits_{x \to -\infty} x^{n}=\infty$ Thus, $S=\dfrac{1}{9}$
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