Q. 18 - Calculus

Question

If $\underset{k\to \infty }{\lim }\frac{a_k}{b_k} = \infty$ and $\sum _{k=1}^{\infty } a_k$ diverges, explain why we cannot draw any conclusions about the behavior of $\sum _{k=1}^{\infty } b_k$.

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{behavior} \mathrm{of} \mathrm{the} \mathrm{series} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{b}}_{\mathrm{k}}}} \mathrm{cannot} \mathrm{be} \mathrm{determined} \mathrm{when} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = \infty \\ \mathrm{and} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{a}}_{\mathrm{k}}}} \mathrm{is} \mathrm{divergent} \mathrm{holds} \mathrm{because} \mathrm{the} \mathrm{series} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{a}}_{\mathrm{k} }}} \mathrm{may} \mathrm{or} \mathrm{may} \mathrm{not} \\ \mathrm{be} \mathrm{divergent}, \mathrm{even} \mathrm{though} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = \infty \mathrm{holds}. \end{gathered}$$

Step 1. Given information

$\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\infty and \sum _{k=1}^{\infty }{a}_k di\mathrm{verges}$ 

Step 2. Finding the value of limit

Consider the convergent series $\sum _{k=1}^{\infty } a_k = \sum _{k=1}^{\infty } \frac{1}{k}$ and the series $\sum _{k=1}^{\infty } b_k = \frac{1}{k^2}$.

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} \mathrm{is}: \\ \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = \underset{\mathrm{k}\to \infty }{\lim } \frac{{\displaystyle \frac{1}{\mathrm{k}}}}{{\displaystyle \frac{1}{{\mathrm{k}}^2}}} \\ = \underset{\mathrm{k}\to \infty }{\lim } \mathrm{k} \\ = \infty \end{gathered}$$

Step 3. Result

$$\begin{gathered} \mathrm{The} \mathrm{series} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{b}}_{\mathrm{k}}}} = \frac{1}{{\mathrm{k}}^2} \mathrm{is} \mathrm{convergent} \mathrm{but} \mathrm{the} \mathrm{series} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{a}}_{\mathrm{k}}}} = \frac{1}{\mathrm{k}} \mathrm{is} \mathrm{divergent}. \\ \mathrm{The} \mathrm{behavior} \mathrm{of} \mathrm{the} \mathrm{series} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{b}}_{\mathrm{k}}}} \mathrm{cannot} \mathrm{be} \mathrm{determined} \mathrm{when} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = \infty \\ \mathrm{and} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{a}}_{\mathrm{k}}}} \mathrm{is} \mathrm{divergent} \mathrm{holds} \mathrm{because} \mathrm{the} \mathrm{series} \underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{a}}_{\mathrm{k} }}} \mathrm{may} \mathrm{or} \mathrm{may} \mathrm{not} \mathrm{be} \\ \mathrm{divergent}, \mathrm{even} \mathrm{though} \underset{\mathrm{k}\to \infty }{\lim } \frac{{\mathrm{a}}_{\mathrm{k}}}{{\mathrm{b}}_{\mathrm{k}}} = \infty \mathrm{holds}. \end{gathered}$$