Q. 46 - Calculus

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.

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

Step-by-Step Solution

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):

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

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 S 10

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

Part (c) Step 1. Finding bound on R 10

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

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

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