Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.6 - Absolute Convergence and the Ratio and Root Tests - 11.6 Exercises - Page 743: 44


Hence, the series converges for $k\geq 2$

Work Step by Step

$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\dfrac{(n+1!)^{2}}{(k(n+1))!}}{\dfrac{(n!)^{2}}{(kn)!}}|$ $=\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(kn+1)(kn+2)....(kn+k)}$ When $k=1$ $\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(kn+1)(kn+2)....(kn+k)}=\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(n+1)} =\infty$ Thus, the series diverges when $k=1$ When $k=2$ $\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(kn+1)(kn+2)....(kn+k)}=\lim\limits_{n \to \infty}\frac{(n+1)^{2}}{(2n+1)(2n+2)} =\frac{1}{4}$ Thus, the series converges when $k=2$ When $k\gt 2$, we have a rational function with the numerator having a smaller degree than the denominator. Thus, the limit must be $0$ when $n \to \infty $. Hence, the series converges for $k\geq 2$
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