University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Questions to Guide Your Review - Page 551: 15

Answer

See below.

Work Step by Step

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{p}}=\frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\cdots+\frac{1}{n^{p}}+\cdots\qquad$ which is a p-series By Example 3 in section 3, the p-series converges for $ p\gt1$ and diverges for $p\leq 1.$ The proof was derived by using the Integral Text. Examples: When $p=1$, we have the harmonic series, $1+\displaystyle \frac{1}{2}+\frac{1}{3}++\frac{1}{n}+...$ which we know from before diverges. When $p=2$, we have the series: $1+\displaystyle \frac{1}{4}+\frac{1}{9}++\frac{1}{n^{2}}+$ which converges by the Integral Test.
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