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: 22



Work Step by Step

Consider $s_n=\dfrac{-8}{(4n-3)(4n+1)}$ Re-write the given series as: $s_n=\dfrac{-8}{(4n-3)(4n+1)}=\dfrac{-2}{(4n-3))}+\dfrac{2}{(4n+1)}$ $s_n=(\dfrac{-2}{9}+\dfrac{2}{13})+(\dfrac{-2}{13}+\dfrac{2}{17})+(\dfrac{-2}{17}+\dfrac{2}{21})+(\dfrac{-2}{21}+\dfrac{-2}{25})...+(\dfrac{-2}{4n-3}+\dfrac{2}{4n+1})$ or, $s_n=\dfrac{-2}{9}+\dfrac{2}{4n+1}$ Applying limits, we get: $\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(\dfrac{-2}{9}+\dfrac{2}{4n+1})$ After simplifications, we get $\lim\limits_{n \to \infty}s_n=(\dfrac{-2}{9}+0)$ Hence, $\lim\limits_{n \to \infty}s_n=\dfrac{-2}{9}$
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