Calculus (3rd Edition)

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

Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 556: 6



Work Step by Step

Given $$\sum_{n=1}^{\infty} \frac{n }{(n^2+1)^{3/5}}$$ Since $f(n)= \frac{n }{(n^2+1)^{3/5}},\ \ f'(n) = \frac{-n^2+5}{5\left(n^2+1\right)^{\frac{8}{5}}} $, is positive, decreasing and continuous for $n\geq1$, then by using the integral test \begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{n }{(n^2+1)^{3/5}}d n \\ &=\left.\lim _{t \rightarrow \infty}\frac{ 5}{4}(n^2+1)^{2/5}\right|_{1} ^{t} \\ &= \lim _{t \rightarrow \infty} \frac{ 5}{4}(t^2+1)^{2/5}-\lim _{t \rightarrow \infty} \frac{ 5}{4}(2)^{2/5} \\ &=\infty\end{aligned} Thus the series diverges.
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