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 - Practice Exercises - Page 636: 20



Work Step by Step

Consider $s_n=\dfrac{-2}{n(n+1)}$ Re-write the given series as: $s_n=\dfrac{-2}{n(n+1)}=(\dfrac{-2}{2}+\dfrac{2}{3})+(\dfrac{-2}{3}+\dfrac{2}{4})+(\dfrac{-2}{4}+\dfrac{2}{5})+(\dfrac{-2}{5}+\dfrac{2}{6})...+(\dfrac{-2}{n}+\dfrac{2}{n+1})$ or, $s_n=\dfrac{-2}{2}+\dfrac{2}{n+1}$ Applying limits, we get: $\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}\dfrac{-2}{2}+\dfrac{2}{n+1}$ After simplifications, we get $\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(-1+\dfrac{2}{n+1})$ $\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(-1+\dfrac{2}{\infty})$ $\lim\limits_{n \to \infty}s_n=-1$
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