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 498: 41


Converges to $1$

Work Step by Step

We have $a_n=\dfrac{4}{(4n-3)(4n+1)}=\dfrac{1}{4n-3}-\dfrac{1}{4n+1}$ Here, $ s_n=(1-\dfrac{1}{5})(\dfrac{1}{5}-\dfrac{1}{9})+....\dfrac{1}{4n-3}-\dfrac{1}{4n+1}=1-\dfrac{1}{4n+1}$ Thus, we have: $\lim\limits_{n \to \infty} s_n=1-\lim\limits_{n \to \infty} \dfrac{1}{4n+1}=1+0=1$ We see that $s_n \to 1$ as $n \to \infty$, so, the given series converges to $1$.
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