Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.4 Absolute and Conditional Convergence - Exercises - Page 563: 4


The series $\sum_{n=1}^{\infty} \frac{(-1)^{n}n^4}{ n^3+1}$ diverges

Work Step by Step

We have the absolute series $$\sum_{n=1}^{\infty} |\frac{(-1)^{n}n^4}{ n^3+1}|=\sum_{n=1}^{\infty} \frac{n^4}{ n^3+1}.$$ So, we have $$\lim_{n\to \infty }b_n=\lim_{n\to \infty } \frac{n^4}{ n^3+1}=\lim_{n\to \infty } \frac{n }{ (1/n)+(1/n^4)}=\infty\neq 0.$$ Then the positive series diverges by the divergence test. The terms $\frac{n^4}{ n^3+1}$ do not form a decreasing sequence. Thus, the series $\sum_{n=1}^{\infty} \frac{(-1)^{n}n^4}{ n^3+1}$ does not converge conditionally or absolutely.
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