Question

For each series in Exercises 44–47, do each of the following:

(a) Use the integral test to show that the series converges.

(b) Use the 10th term in the sequence of partial sums to approximate the sum of the series.

(c) Use Theorem 7.31 to find a bound on the tenth remainder $R_{10}$.

(d) Use your answers from parts (b) and (c) to find an interval containing the sum of the series.

(e) Find the smallest value of n so that.

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

Short answer

Part (a):

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

Part (b):

$$\begin{gathered} {\mathrm{S}}_{10}=\frac{1}{2{\left(\ln \left(2\right)\right)}^2}+\frac{1}{3{\left(\ln \left(3\right)\right)}^2}+\frac{1}{4{\left(\ln \left(4\right)\right)}^2}+...+\frac{1}{11{\left(\ln \left(11\right)\right)}^2} \\ =1.0406+0.2762+0.13+0.0772+...+0.0158\approx 1.7002 \\ \mathrm{Therefore},\mathrm{the}{10}^{\mathrm{th}}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{sequence}\mathrm{of}\mathrm{partial}\mathrm{sums}\mathrm{to}\mathrm{approximate} \\ \mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{series}\mathrm{is}{\mathrm{S}}_{10}\approx 1.7002 \end{gathered}$$

Part (c):

$$\begin{gathered} T\mathrm{he}\mathrm{tenth}\mathrm{remainder}{\mathrm{R}}_{10},\mathrm{is}\mathrm{bounded}\mathrm{by}: \\ 0\le {\mathrm{R}}_{10}\le {\int }_{\mathrm{x}=10}^{\infty }\frac{1}{x{\left(\ln \left(x\right)\right)}^2}\mathrm{dx}.....\left(1\right) \\ {\int }_{\mathrm{x}=10}^{\infty }\frac{1}{x{\left(\ln \left(x\right)\right)}^2}\mathrm{dx}\approx 0.43429 \\ \mathrm{Therefore}: \\ 0\le {\mathrm{R}}_{10}\le 0.43429 \\ \mathrm{Therefore},\mathrm{the}\mathrm{bound}\mathrm{on}{10}^{\mathrm{th}}\mathrm{remainder},{\mathrm{R}}_{10}\mathrm{is}0.43429 \end{gathered}$$

Part (d):

$$\begin{gathered} {\mathrm{S}}_{10}\le \mathrm{L}\le {\mathrm{S}}_{10}+{\mathrm{B}}_{10} \\ 1.7002\le \mathrm{L}\le 1.7002+0.43429\left(\mathrm{Substitution}\right) \\ 1.7002\le \mathrm{L}\le 2.13449 \\ \mathrm{Therefore},\mathrm{the}\mathrm{interval}\mathrm{containing}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{series}\mathrm{is}\mathrm{L}\in (1.7002,2.13449). \end{gathered}$$

Part (e):

$$\begin{gathered} {\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx}\le {10}^{-6} \\ \frac{1}{\ln \left(n\right)}\le {10}^{-6} \\ \ln \left(n\right)\ge 1000000 \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{n}\mathrm{so}\mathrm{that}{\mathrm{R}}_{\mathrm{n}}\le {10}^{-6}\mathrm{holds}\mathrm{is}\mathrm{n}=3x{10}^{434294} \end{gathered}$$

Step-by-step solution

Step 1. Given information is:

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

Part (a) Step 1. Examining nature of given function: 

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^2}. \\ \mathrm{The}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^2}\mathrm{is}\mathrm{continuous},\mathrm{decreasing},\mathrm{with}\mathrm{positive}\mathrm{terms}. \\ \mathrm{Therefore}\mathrm{all}\mathrm{the}\mathrm{conditions}\mathrm{of}\mathrm{integral}\mathrm{test}\mathrm{are}\mathrm{fulfilled}. \\ \mathrm{So},\mathrm{integral}\mathrm{test}\mathrm{is}\mathrm{applicable}. \end{gathered}$$

Part (a) Step 2. Solving the integral: 

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{integral}:{\int }_{\mathrm{x}=2}^{\infty }\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{x}=2}^{\infty }\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^2}\mathrm{dx}. \\ \mathrm{Therefore}, \\ {\int }_{\mathrm{x}=2}^{\infty }\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^2}\mathrm{dx}=\underset{\mathrm{k}\to \infty }{\lim }{\int }_{\mathrm{x}=2}^{\mathrm{k}}\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^2}\mathrm{dx} \\ =\underset{\mathrm{k}\to \infty }{\lim }{\int }_{\mathrm{u}=\ln \left(2\right)}^{\ln \left(\mathrm{k}\right)}\frac{1}{{\mathrm{u}}^2}\mathrm{du}\left(\mathrm{Put}\ln \left(\mathrm{x}\right)=\mathrm{u},\Rightarrow \frac{1}{\mathrm{x}}\mathrm{dx}=\mathrm{du}\right) \\ =\underset{\mathrm{k}\to \infty }{\lim }{\left[-\frac{1}{\mathrm{u}}\right]}_{\ln \left(2\right)}^{\ln \left(\mathrm{k}\right)} \\ =\underset{\mathrm{k}\to \infty }{\lim }\left[-\frac{1}{\ln \left(\mathrm{k}\right)}+\frac{1}{\ln \left(2\right)}\right]\left(\mathrm{Substitution}\right) \\ =0+\frac{1}{\ln \left(2\right)} \\ =\frac{1}{\ln \left(2\right)} \end{gathered}$$

Part (a) Step 3. Result

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

Part (b) Step 1. Finding S10

$$\begin{gathered} \mathrm{The}\mathrm{value}\mathrm{of}{\mathrm{S}}_{10}\mathrm{is}: \\ {\mathrm{S}}_{10}=\frac{1}{2{\left(\ln \left(2\right)\right)}^2}+\frac{1}{3{\left(\ln \left(3\right)\right)}^2}+\frac{1}{4{\left(\ln \left(4\right)\right)}^2}+...+\frac{1}{11{\left(\ln \left(11\right)\right)}^2} \\ =1.0406+0.2762+0.13+0.0772+...+0.0158\approx 1.7002 \\ \mathrm{Therefore},\mathrm{the}{10}^{\mathrm{th}}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{sequence}\mathrm{of}\mathrm{partial}\mathrm{sums}\mathrm{to}\mathrm{approximate} \\ \mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{series}\mathrm{is}{\mathrm{S}}_{10}\approx 1.7002 \end{gathered}$$

Part (c) Step 1. Finding bound on R10

$$\begin{gathered} \mathrm{If}\mathrm{a}\mathrm{function}\mathrm{a}\mathrm{is}\mathrm{continuous},\mathrm{positive},\mathrm{and}\mathrm{decreasing},\mathrm{and}\mathrm{if}\mathrm{the}\mathrm{improper} \\ \mathrm{Integral}{\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{converges},\mathrm{then}\mathrm{the}\mathrm{nth}\mathrm{remainder},{\mathrm{R}}_{\mathrm{n}},\mathrm{for}\mathrm{the}\mathrm{series} \\ \sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right)\mathrm{is}\mathrm{bounded}\mathrm{by}: \\ 0\le {\mathrm{R}}_{\mathrm{n}}\le {\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} \\ \mathrm{Furthermore},\mathrm{if}{\mathrm{B}}_{\mathrm{n}}={\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx},\mathrm{then} \\ {\mathrm{S}}_{\mathrm{n}}\le \sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right)\le {\mathrm{S}}_{\mathrm{n}}+{\mathrm{B}}_{\mathrm{n}} \\ \mathrm{The}\mathrm{result}\mathrm{gives}\mathrm{that}\mathrm{the}\mathrm{tenth}\mathrm{remainder}{\mathrm{R}}_{10},\mathrm{is}\mathrm{bounded}\mathrm{by}: \\ 0\le {\mathrm{R}}_{10}\le {\int }_{\mathrm{x}=10}^{\infty }\frac{1}{x{\left(\ln \left(x\right)\right)}^2}\mathrm{dx}.....\left(1\right) \\ \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{integral}{\int }_{\mathrm{x}=10}^{\infty }\frac{1}{\mathrm{x}{\left(\ln \left(\mathrm{x}\right)\right)}^2}\mathrm{dx}\mathrm{is}: \\ {\int }_{\mathrm{x}=10}^{\infty }\frac{1}{x{\left(\ln \left(x\right)\right)}^2}\mathrm{dx}\approx 0.43429(\mathrm{Value}\mathrm{of}\mathrm{integral}\mathrm{solved}\mathrm{above}\mathrm{in}(\mathrm{a}\left)\right) \\ \mathrm{Therefore},\mathrm{equation}\left(1\right)\mathrm{becomes}: \\ 0\le {\mathrm{R}}_{10}\le 0.43429 \\ \mathrm{Therefore},\mathrm{the}\mathrm{bound}\mathrm{on}{10}^{\mathrm{th}}\mathrm{remainder},{\mathrm{R}}_{10}\mathrm{is}0.43429 \end{gathered}$$

Part (d) Step 1. Finding Interval

$$\begin{gathered} \mathrm{Assume}{\mathrm{B}}_{10}\approx 0.43429\mathrm{and}\sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right)=\mathrm{L}. \\ \mathrm{Using}\mathrm{the}\mathrm{result}{\mathrm{S}}_{\mathrm{n}}\le \sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right)\le {\mathrm{S}}_{\mathrm{n}}+{\mathrm{B}}_{\mathrm{n}},\mathrm{the}\mathrm{following}\mathrm{inequality}\mathrm{is}\mathrm{obtained}. \\ {\mathrm{S}}_{10}\le \mathrm{L}\le {\mathrm{S}}_{10}+{\mathrm{B}}_{10} \\ 1.7002\le \mathrm{L}\le 1.7002+0.43429\left(\mathrm{Substitution}\right) \\ 1.7002\le \mathrm{L}\le 2.13449 \\ \mathrm{Therefore},\mathrm{the}\mathrm{interval}\mathrm{containing}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{series}\mathrm{is}\mathrm{L}\in (1.7002,2.13449). \end{gathered}$$

Part (e) Step 1. Finding n

$$\begin{gathered} \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{smallest}\mathrm{value}\mathrm{of}\mathrm{n},\mathrm{find}\mathrm{n}\mathrm{such}\mathrm{that}\mathrm{following}\mathrm{holds}: \\ {\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx}\le {10}^{-6} \\ \frac{1}{\ln \left(n\right)}\le {10}^{-6} \\ \ln \left(n\right)\ge 1000000 \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{n}\mathrm{so}\mathrm{that}{\mathrm{R}}_{\mathrm{n}}\le {10}^{-6}\mathrm{holds}\mathrm{is}\mathrm{n}=3x{10}^{434294} \end{gathered}$$