Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 638: 28

\[\text{The series converges}\]

\[\begin{align} & \sum\limits_{k=1}^{\infty }{\frac{k}{{{\left( {{k}^{2}}+1 \right)}^{3}}}} \\ & \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=\frac{k}{{{\left( {{k}^{2}}+1 \right)}^{3}}},\text{ then let }f\left( x \right)=\frac{x}{{{\left( {{x}^{2}}+1 \right)}^{3}}} \\ & \text{The domain of }f\left( x \right)\text{is }\left( -\infty ,\infty \right)\text{, so }f\left( x \right)\,\text{is continuous and positive } \\ & \text{for }x\ge 1, \\ & \text{Now we will determine if }f\left( x \right)\text{ is decreasing}\text{, then } \\ & f\left( x \right)=\frac{x}{{{\left( {{x}^{2}}+1 \right)}^{3}}} \\ & \text{Differentiating} \\ & f'\left( x \right)=\frac{{{\left( {{x}^{2}}+1 \right)}^{3}}-3x{{\left( {{x}^{2}}+1 \right)}^{2}}\left( 2x \right)}{{{\left( {{x}^{2}}+1 \right)}^{6}}} \\ & f'\left( x \right)=\frac{\left( {{x}^{2}}+1 \right)-3x\left( 2x \right)}{{{\left( {{x}^{2}}+1 \right)}^{4}}} \\ & f'\left( x \right)=\frac{1-6{{x}^{2}}}{{{\left( {{x}^{2}}+1 \right)}^{4}}} \\ & \text{Critical points }x=\pm \frac{1}{\sqrt{6}} \\ & \text{Test the interval }\left( 1,\infty \right) \\ & f'\left( 2 \right)=\frac{1-6{{\left( 2 \right)}^{2}}}{{{\left( {{2}^{2}}+1 \right)}^{4}}}<0,\text{ decreasing} \\ & \text{The function satisfies the conditions for the integral test}\text{.} \\ & \\ & \int_{1}^{\infty }{\frac{x}{{{\left( {{x}^{2}}+1 \right)}^{3}}}}dx \\ & \text{Solving improper integral} \\ & \int_{1}^{\infty }{\frac{x}{{{\left( {{x}^{2}}+1 \right)}^{3}}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{\frac{x}{{{\left( {{x}^{2}}+1 \right)}^{3}}}}dx \\ & \text{ }=\frac{1}{2}\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{{{\left( {{x}^{2}}+1 \right)}^{-3}}\left( 2x \right)}dx \\ & \text{Integrating}\text{, we obtain} \\ & \text{ }=\frac{1}{2}\underset{b\to \infty }{\mathop{\lim }}\,\left[ \frac{{{\left( {{x}^{2}}+1 \right)}^{-2}}}{-2} \right]_{1}^{b} \\ & \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{\left( {{b}^{2}}+1 \right)}-\frac{1}{\left( {{1}^{2}}+1 \right)} \right] \\ & \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{\left( {{b}^{2}}+1 \right)} \right]-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{2} \right] \\ & \text{Evaluate the limit when }b\to \infty \\ & \text{ }=0-\frac{1}{8} \\ & \text{ }=-\frac{1}{8} \\ & \text{The integral converges}\text{, so by the Integral test }\left( \text{Theorem 8}\text{.10} \right) \\ & \text{The series converges} \\ \end{align}\]