Chapter 8 - Sequences and Infinite Series - Review Exercises - Page 659: 15

\[\text{The series converges}\]

\[\begin{align} & \sum\limits_{k=1}^{\infty }{\frac{1}{k\left( k+1 \right)}} \\ & \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=\frac{1}{k\left( k+1 \right)},\text{ then let }f\left( x \right)=\frac{1}{x\left( x+1 \right)} \\ & \text{Then }f\left( x \right)\,\text{is continuous and positive } \\ & \text{for }x\ge 1, \\ & \text{This function is decreasing for }x\ge 1,\text{ then} \\ & \text{The function satisfies the conditions for the integral test}\text{.} \\ & \\ & \int_{1}^{\infty }{\frac{1}{x\left( x+1 \right)}dx} \\ & \text{Solving the improper integral} \\ & \int_{1}^{\infty }{\frac{1}{x\left( x+1 \right)}dx}=\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{\frac{1}{x\left( x+1 \right)}}dx \\ & \text{*Decomposing the integrand into partial fractions} \\ & \frac{1}{x\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x+1} \\ & 1=A\left( x+1 \right)+Bx \\ & x=0\to A=1 \\ & x=-1\to B=-1 \\ & \frac{1}{x\left( x+1 \right)}=\frac{1}{x}-\frac{1}{x+1} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{\left[ \frac{1}{x}-\frac{1}{x+1} \right]}dx \\ & \text{Integrating} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| x \right|-\ln \left| x+1 \right| \right]_{1}^{b} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| \frac{x}{x+1} \right| \right]_{1}^{b} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| 1-\frac{1}{x+1} \right| \right]_{1}^{b} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| 1-\frac{1}{b+1} \right|-\ln \left| 1-\frac{1}{1+1} \right| \right] \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| 1-\frac{1}{b+1} \right|-\ln \left| \frac{1}{2} \right| \right] \\ & \text{Evaluate the limit when }b\to \infty \\ & =\ln \left| 1-0 \right|-\ln \left| \frac{1}{2} \right| \\ & =-\ln \left| \frac{1}{2} \right| \\ & \text{The integral converges}\text{, so by the Integral test }\left( \text{Theorem 8}\text{.10} \right) \\ & \text{The series converges} \\ \end{align}\]