Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.6 - Absolute Convergence and the Ratio and Root Tests - 11.6 Exercises: 8

Answer

The series converges.

Work Step by Step

The ratio test states that, for an infinite series, if $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|<1$, the series is convergent. If $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|>1$ or $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty$, the series is divergent. If $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=0$, we cannot make a conclusion. We use this test to determine whether the series $$\sum_{n=1}^{\infty}\frac{(-2)^n}{n^2}$$ is convergent or divergent. First, we set up the limit according to the formula $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$ by plugging in $n+1$ for $n$ in the numerator and leaving $n$ alone in the denominator: $$\lim\limits_{n \to \infty}\left|\frac{(-2)^{n+1}}{n^{2+1}}\div\frac{(-2)^n}{n^2}\right|$$We simplify the division of fractions by multiplying by the reciprocal: $$\lim\limits_{n \to \infty}\left|\frac{n^2(-2)^{n+1}}{(-2)^nn^3}\right|$$We simplify by subtracting exponents: $$\lim\limits_{n \to \infty}\left|\frac{-2}{n}\right|$$The limit as $n\to\infty$ of a fraction whose denominator has a greater power than its numerator is $0$. In this case, $$\lim\limits_{n \to \infty}\left|\frac{-2}{n}\right|=0$$Since $0<1$, the series converges.
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