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: 28



Work Step by Step

We have: $\Sigma_{n=1}^\infty \dfrac{n(n+1)}{(n+2)(n+3)}$ and $\lim\limits_{n \to \infty} \dfrac{n(n+1)}{(n+2)(n+3)}=\lim\limits_{n \to \infty} \dfrac{n^2(1+\dfrac{1}{n})}{n(1+\dfrac{2}{n})(1+\dfrac{3}{n})}$ Now, $\lim\limits_{n \to \infty} \dfrac{n^2(1+\dfrac{1}{n})}{n(1+\dfrac{2}{n})(1+\dfrac{3}{n})}=\dfrac{\lim\limits_{n \to \infty} 1+\lim\limits_{n \to \infty} \dfrac{1}{n}}{\lim\limits_{n \to \infty} (1+\dfrac{2}{n})+ \lim\limits_{n \to \infty} (1+\dfrac{3}{n}){}}=\dfrac{1+0}{1 \cdot 1}=1$ This shows a divergent series in accordance to the nth-Term Integral Test.
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