Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 639: 53

\[\text{The series converges}\]

\[\begin{align} & \sum\limits_{k=1}^{\infty }{\frac{1}{\left( 3k+1 \right)\left( 3k+4 \right)}} \\ & \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=\frac{1}{\left( 3k+1 \right)\left( 3k+4 \right)},\text{ then let }f\left( x \right)=\frac{1}{\left( 3x+1 \right)\left( 3x+4 \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}{\left( 3x+1 \right)\left( 3x+4 \right)}dx} \\ & \text{Solving the improper integral} \\ & \int_{1}^{\infty }{\frac{1}{\left( 3x+1 \right)\left( 3x+4 \right)}dx}=\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{\frac{1}{\left( 3x+1 \right)\left( 3x+4 \right)}}dx \\ & \text{*Decomposing the integrand into partial fractions} \\ & \frac{1}{\left( 3x+1 \right)\left( 3x+4 \right)}=\frac{A}{3x+1}+\frac{B}{3x+4} \\ & 1=A\left( 3x+4 \right)+B\left( 3x+1 \right) \\ & x=-\frac{1}{3}\to A=3 \\ & x=-\frac{4}{3}\to B=-3 \\ & \frac{1}{\left( 3x+1 \right)\left( 3x+4 \right)}=\frac{3}{3x+1}-\frac{3}{3x+4} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{\left[ \frac{3}{3x+1}-\frac{3}{3x+4} \right]}dx \\ & \text{Integrating} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| 3x+1 \right|-\ln \left| 3x+4 \right| \right]_{1}^{b} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| \frac{3x+1}{3x+4} \right| \right]_{1}^{b} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| 1-\frac{3}{3x+4} \right| \right]_{1}^{b} \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| 1-\frac{3}{3b+4} \right|-\ln \left| 1-\frac{3}{3\left( 1 \right)+4} \right| \right] \\ & =\underset{b\to \infty }{\mathop{\lim }}\,\left[ \ln \left| 1-\frac{3}{3b+4} \right|-\ln \left| \frac{4}{7} \right| \right] \\ & \text{Evaluate the limit when }b\to \infty \\ & =\ln \left| 1-0 \right|-\ln \left| \frac{4}{7} \right| \\ & =-\ln \left| \frac{4}{7} \right| \\ & \text{The integral converges}\text{, so by the Integral test }\left( \text{Theorem 8}\text{.10} \right) \\ & \text{The series converges} \\ \end{align}\]