Q. 49 - Calculus

Question

In Exercises 48–51 find all values of p so that the series converges.

$\sum _{k=2}^{\infty }\frac{1}{k{\left(\ln \left(k\right)\right)}^p}$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{integral} {\int }_{\mathrm{x}=2}^{\infty }\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^{\mathrm{p}}}\mathrm{dx} \mathrm{converges} \mathrm{for} \mathrm{p}>1. \\ \mathrm{Thus}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=2}^{\infty }\frac{1}{\mathrm{k}{\left(\ln \left(\mathrm{k}\right)\right)}^{\mathrm{p}}} \mathrm{is} \mathrm{convergent} \mathrm{for} \mathrm{p}>1. \end{gathered}$$

Step 1. Given information:

$\sum _{k=2}^{\infty }\frac{1}{k{\left(\ln \left(k\right)\right)}^p}$

Step 2. Examining nature of given function:

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) = \frac{1}{x{\left(\ln \left(x\right)\right)}^p}. \\ \mathrm{The} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) = \frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^{\mathrm{p}}} \mathrm{is} \mathrm{continuous}, \mathrm{decreasing}, \mathrm{with} \mathrm{positive} \mathrm{terms}. \\ \mathrm{Therefore} \mathrm{all} \mathrm{the} \mathrm{conditions} \mathrm{of} \mathrm{integral} \mathrm{test} \mathrm{are} \mathrm{fulfilled}. \\ \mathrm{So}, \mathrm{integral} \mathrm{test} \mathrm{is} \mathrm{applicable}. \end{gathered}$$

Step 3. Solving the integral:

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{integral}: {\int }_{\mathrm{x}=2}^{\infty } \mathrm{f}\left(\mathrm{x}\right) \mathrm{dx} = {\int }_{\mathrm{x}=2}^{\infty }\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^{\mathrm{p}}}\mathrm{dx}. \\ \mathrm{Therefore}, \\ {\int }_{\mathrm{x}=2}^{\infty }\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^{\mathrm{p}}} \mathrm{dx} = \underset{\mathrm{k}\to \infty }{\lim }{\int }_{\mathrm{x}=2}^{\mathrm{k}}\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^{\mathrm{p}}}\mathrm{dx} \\ =\underset{\mathrm{k}\to \infty }{\lim }{\int }_{\mathrm{u}=\ln \left(2\right)}^{\ln \left(\mathrm{k}\right)}\frac{1}{{\mathrm{u}}^{\mathrm{p}}}\mathrm{du} \left(\mathrm{Put} \ln \left(\mathrm{x}\right) = \mathrm{u}, \Rightarrow \frac{1}{\mathrm{x}}\mathrm{dx} = \mathrm{du}\right) \\ = \underset{\mathrm{k}\to \infty }{\lim }{\left[\frac{1}{\left(-\mathrm{p}+1\right) {\mathrm{u}}^{\mathrm{p}-1}}\right]}_{\ln \left(2\right)}^{\ln \left(\mathrm{k}\right)} \\ = \frac{1}{\left(-\mathrm{p}+1\right)} \underset{\mathrm{k}\to \infty }{\lim }{\left[\frac{1}{{\left(\ln \left(\mathrm{k}\right)\right)}^{\mathrm{p}-1}}-\frac{1}{\ln \left(2\right)}\right]}_{\ln \left(2\right)}^{\ln \left(\mathrm{k}\right)} \left(\mathrm{Substitution}\right) \end{gathered}$$

Step 4. Result:

$$\begin{gathered} \mathrm{The} \mathrm{improper} \mathrm{integral} \mathrm{converges} \mathrm{to} \mathrm{finite} \mathrm{value} \mathrm{only} \mathrm{when} \mathrm{p}>1. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{integral} {\int }_{\mathrm{x}=2}^{\infty }\frac{1}{x{\left(\ln \left(x\right)\right)}^p}\mathrm{dx} \mathrm{converges} \mathrm{for} \mathrm{p}>1. \\ \mathrm{Thus}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=2}^{\infty }\frac{1}{\mathrm{k}{\left(\ln \left(\mathrm{k}\right)\right)}^{\mathrm{p}}} \mathrm{is} \mathrm{convergent} \mathrm{for} \mathrm{p}>1. \end{gathered}$$