Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.4 The Comparison Tests - 11.4 Exercises - Page 771: 4



Work Step by Step

Given $$\sum_{n=2}^{\infty} \frac{1 }{\sqrt{n}-1}$$ Use the Limit Comparison Test with $a_n =\dfrac{1 }{\sqrt{n}-1}$ and $b_n=\dfrac{1 }{\sqrt{n} }$ \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{\sqrt{n} }{\sqrt{n}-1}\\ &=1 \end{align*} Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ is divergent ($p-$ series $p<1$) , then $\displaystyle\sum_{n=2}^{\infty}\frac{1}{\sqrt{n}-1}$ also divergent
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