Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 766: 10



Work Step by Step

The p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\sum_{n=3}^{\infty}n^{-0.9999}=\sum_{n=3}^{\infty}\frac{1}{n^{0.9999}}$ The series $\sum_{n=1}^{\infty}\frac{1}{n^{0.9999}}$is a p-series with $p= 0.9999 \leq 1$ and it is divergent. But the given series is not exactly a p-series, it is because the summation starts from $n=3$ But adding or subtracting finite number of terms will not change the convergence/divergence nature of the series. Hence, the given series is divergent.
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