Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 636: 26



Work Step by Step

A $p$-series has the form of $\Sigma_{n=k}^{\infty}\dfrac{1}{n^p}$.It is convergent iff $p \gt 1$ otherwise diverges. From the given problem, we have $\Sigma_{n=1}^{\infty}\dfrac{-5}{n}$ The given series can be re-written as: $a_n=-5\Sigma_{n=1}^{\infty}\dfrac{1}{ n^{1}}$ Here, $p=1$ Thus, the given series divergent.
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