Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 9 - Infinite Series - 9.3 Exercises - Page 609: 21

Answer

$$ \sum_{n=1}^{\infty} \frac{n}{n^{4}+1} $$ Apply the Integral Test to the given series we obtain $$ \int_{1}^{\infty}\frac{x}{x^{4}+1} d x =\frac{\pi}{8}. $$ So, the series converges

Work Step by Step

$$ \sum_{n=1}^{\infty} \frac{n}{n^{4}+1} $$ Apply the Integral Test to the series $$ \sum_{n=1}^{\infty} \frac{n}{n^{4}+1} $$ The function $$ f(x)=\frac{x}{x^{4}+1} $$ is positive and continuous for $x\geq 1$ . To determine whether $f$ is decreasing, find the derivative. $$ f^{'}(x)=\frac{1-3x^{4}} {(x^{4}+1)^{2}} $$ So, $f^{'}(x) \lt 0 $ for $x \geq 1$, and it follows that $f$ satisfies the conditions for the Integral Test. You can integrate to obtain $$ \begin{aligned} \int_{1}^{\infty}\frac{x}{x^{4}+1} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{x}{x^{4}+1} d x \\ & \quad\quad\quad \left[\text { use integration by substituting }\right] \\ &\quad\quad\quad \left[ u=x^{2}, \quad\quad d u= 2x dx \right] , \text { then }\\ &= \lim\limits_{t \to \infty } \frac{1}{2}\int_{1}^{t } \frac{2x}{x^{4}+1} d x \\ &= \lim\limits_{t \to \infty } \frac{1}{2}\int_{1}^{t } \frac{du}{u^{2}+1} \\ &=\lim\limits_{t \to \infty } \left[ \frac{1}{2 }\arctan u \right] _{1}^{t} \\ &=\lim\limits_{t \to \infty } \left[ \frac{1}{2 }\arctan x^{2} \right] _{1}^{t} \\ &= \frac{1}{2 } \left[ \lim\limits_{t \to \infty } \arctan t^{2}-\frac{\pi}{4} \right] \\ &= \frac{1}{2 } \left[ \frac{\pi}{2}-\frac{\pi}{4} \right] \\ &=\frac{\pi}{8}. \end{aligned} $$ So, the series converges.
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