## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 90

#### Answer

Converges to $\dfrac{1}{p-1}$ when $p \gt 1$

#### Work Step by Step

Since, we have $\int_1^n \dfrac{1}{x^p} dx=[\dfrac{1}{-p+1}(x^{-p+1}]_1^n=\dfrac{1}{1-p}(\dfrac{1}{n^{p-1}-1})$ Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{1-p}(\dfrac{1}{n^{p-1}-1})$ Now, if $p \lt 1$ then $\dfrac{1}{n^{p-1}-1}$ approaches to $\infty$ if $p=1$ then we have $\dfrac{1}{1-p}(\dfrac{1}{n^{p-1}-1}) \to \dfrac{1}{0}$ if $p \gt 1$ then we have $\dfrac{1}{1-p}(\dfrac{1}{n^{p-1}-1}) \to \dfrac{1}{p-1}$ Hence, $\lim\limits_{n \to \infty} a_n=\dfrac{1}{p-1}$ and {$a_n$} is Convergent and converges to $\dfrac{1}{p-1}$ when $p \gt 1$

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