Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 638: 22

\[\text{The series diverges}\]

\[\begin{align} & \sum\limits_{k=1}^{\infty }{\frac{1}{\sqrt[3]{k+10}}} \\ & \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=\frac{1}{\sqrt[3]{k+10}},\text{ then let }f\left( x \right)=\frac{1}{\sqrt[3]{x+10}} \\ & \text{ }f\left( x \right)=\frac{1}{\sqrt[3]{x+10}}\text{ is continuous and positive for }x\ge 1, \\ & \text{now we will determine if }f\left( x \right)\text{ is decreasing on }\left( 1,\infty \right)\text{, then } \\ & f\left( x \right)=\frac{1}{\sqrt[3]{x+10}} \\ & f\left( x \right)={{\left( x+10 \right)}^{-1/3}} \\ & \text{Differentiating } \\ & f'\left( x \right)=-\frac{1}{3}{{\left( x+10 \right)}^{-4/3}} \\ & \text{Critical point at }x=-10 \\ & \\ & \text{Intervals }\left( -\infty ,-10 \right)\text{ and }\left( -10,\infty \right) \\ & \text{*Test for }\left( -\infty ,-10 \right) \\ & f'\left( -11 \right)=-\frac{1}{3}{{\left( -11+10 \right)}^{-4/3}}=-\frac{1}{3}<0 \\ & \text{*Test for }\left( -10,\infty \right) \\ & f'\left( -9 \right)=-\frac{1}{3}{{\left( -9+10 \right)}^{-4/3}}=-\frac{1}{3}<0 \\ & \text{Therefore}\text{, the function is decreasing for }\left( 1,\infty \right)\text{ } \\ & \text{The function satisfies the conditions for the integral test}\text{.} \\ & \\ & \int_{1}^{\infty }{\frac{1}{\sqrt[3]{x+10}}}dx \\ & \text{Solving improper integral} \\ & \int_{1}^{\infty }{\frac{1}{\sqrt[3]{x+10}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{{{\left( x+10 \right)}^{-1/3}}}dx \\ & \text{ }=\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{\left( x+10 \right)}^{2/3}} \right]_{1}^{b} \\ & \text{ }=\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{\left( b+10 \right)}^{2/3}} \right]-\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{\left( 11 \right)}^{2/3}} \right] \\ & \text{Evaluate the limit when }b\to \infty \\ & \text{ }=\infty -{{\left( 11 \right)}^{2/3}} \\ & \text{ }=\infty \\ & \text{The integral diverges}\text{, so by the Integral test }\left( \text{Theorem 8}\text{.10} \right) \\ & \text{The series diverges} \\ \end{align}\]