University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.6 - Alternating Series and Conditional Convergence - Exercises - Page 522: 62


Convergent and converge to $1$.

Work Step by Step

We have $_n=\Sigma_{n=1}^{\infty} \dfrac{1}{k (k+1)}=\Sigma_{n=1}^{\infty} \dfrac{1}{k }-\dfrac{1}{{(k+1)}}$ This gives that $s_n=1-\dfrac{1}{n+1}$ Thus, $\lim\limits_{n \to \infty} s_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n+1})=1$ Now, $T_{2n+1}=T_{2n}+\dfrac{1}{n+1}=(1-\dfrac{1}{n+1})+\dfrac{1}{n+1}=1$ Thus, $\lim\limits_{n \to \infty} s_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n+1})=1$ So, both series are convergent and they converge to $1$.
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