University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 497: 36


The series converges to $3$.

Work Step by Step

Here, $ s_n=(\dfrac{3}{1}-\dfrac{3}{4}) +(\dfrac{3}{4}-\dfrac{3}{9})+.....+(\dfrac{3}{n^2}-\dfrac{3}{(n+1)^2})=3-\dfrac{3}{(n+1)^2}$ Thus, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} 3- \lim\limits_{n \to \infty}\dfrac{3}{(n+1)^2}=3$ Now, $s_n \to 3$ as $n \to \infty$, so, the given series will converge to $3$.
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