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

Published by Pearson

# Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 52

Converges

#### Work Step by Step

Consider $a_n=\dfrac{\sqrt[n] n}{n^2}$ and $b_n=\dfrac{1}{ n^2}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{\sqrt[n] n}{n^2}}{\dfrac{1}{ n^2}}$ or, $=\lim\limits_{n \to \infty} \sqrt[n] n$ and $\lim\limits_{n \to \infty} n^{1/n}=1 \ne 0 \ne \infty$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ is a convergent p-series with $p \gt 1$ Thus, the series converges by the limit comparison test.

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