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: 35

Answer

Converges Absolutely

Work Step by Step

Consider $a_n=\dfrac{n+1}{n!}$ By the Ratio Test: $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} |\dfrac{\dfrac{(n+1)+1}{(n+1)!}}{\dfrac{n+1}{n!}}|$ or, $=\lim\limits_{n \to \infty} |\dfrac{\dfrac{n+2}{n!(n+1)!}}{\dfrac{n+1}{n!}}|$ or, $\lim\limits_{n \to \infty} \dfrac{(n+2)}{(n+1)^2}=\lim\limits_{n \to \infty} \dfrac{1}{(n+1)}+\lim\limits_{n \to \infty} \dfrac{1}{(n+1)^2}$ or, $\dfrac{1}{\infty}+\dfrac{1}{\infty}=0$ so, $l=0 \lt 1$ Thus, the series Converges Absolutely by the Ratio Test.
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