University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Practice Exercises - Page 552: 21

Answer

$\dfrac{3}{2}$

Work Step by Step

Consider $s_n=\dfrac{9}{(3n-1)(3n+2)}$ Re-write the given series as: $s_n=\dfrac{9}{(3n-1)(3n+2)}=\dfrac{3}{(3n-1))}-\dfrac{9}{(3n+2)}$ $s_n=(\dfrac{3}{2}-\dfrac{3}{5})+(\dfrac{3}{5}-\dfrac{3}{8})+(\dfrac{3}{8}-\dfrac{3}{11})+(\dfrac{3}{11}-\dfrac{3}{14})...+(\dfrac{3}{3n-1}-\dfrac{3}{3n+2})$ or, $s_n=\dfrac{3}{2}-\dfrac{3}{3n+2}$ Apply limits, we get: $\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(\dfrac{3}{2}-\dfrac{3}{3n+2})$ After simplifications, we get $\lim\limits_{n \to \infty}s_n=(\dfrac{3}{2}-0)$ Hence, $\lim\limits_{n \to \infty}s_n=\dfrac{3}{2}$
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