Question

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

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

Short 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-by-step solution

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:

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