Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 638: 25

\[\text{The series converges}\]

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