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 - Section 10.3 - The Integral Test - Exercises 10.3 - Page 586: 9



Work Step by Step

Let us consider $f(x)=\dfrac{x^2}{e^{x/3}}$ Here, the function $f(x)$ is positive, continuous for $x \geq 1$ and $f(x)$ is decreasing for $x \gt 7$ Then $\int_7^\infty \dfrac{x^2}{e^{x/3}}dx= \lim\limits_{k \to \infty} \int_7^k \dfrac{x^2}{e^{x/3}}dx$ or, $ \lim\limits_{k \to \infty} [\dfrac{-54}{e^{k/3}}+\dfrac{327}{e^{7/3}}]=\dfrac{327}{e^{7/3}}$ Thus, the sequence $\Sigma_{n=1}^\infty \dfrac{n^2}{e^{n/3}}$ is convergent.
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