Q. 45 - 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 $R_{n }\le {10}^{-6}$ .

$\sum _{k=0}^{\infty }{e}^{-k}$

Step-by-Step Solution

Answer

Part (a):

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{integral} \mathrm{is} {\int }_{\mathrm{x}=0}^{\infty }{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}=1 \\ \mathrm{The} \mathrm{integral} \mathrm{converges}. \\ \mathrm{Thus}, \mathrm{the} \mathrm{series} \underset{\mathrm{k}=0}{\overset{\infty }{\sum {\mathrm{e}}^{-k}}} \mathrm{is} \mathrm{convergent}. \end{gathered}$$

Part (b): 

$$\begin{gathered} {\mathrm{S}}_{10} = \frac{1}{e^0}+\frac{1}{e^1} +\frac{1}{e^2}+...+\frac{1}{e^9} \\ \approx 1.589150 \\ \mathrm{Therefore}, \mathrm{the} {10}^{\mathrm{th}} \mathrm{term} \mathrm{in} \mathrm{the} \mathrm{sequence} \mathrm{of} \mathrm{partial} \mathrm{sums} \mathrm{to} \mathrm{approximate} \\ \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{series} \mathrm{is} {\mathrm{S}}_{10} \approx 1.589150 \end{gathered}$$

Part (c): 

$$\begin{gathered} T\mathrm{he} \mathrm{tenth} \mathrm{remainder} {\mathrm{R}}_{10}, \mathrm{is} \mathrm{bounded} \mathrm{by}: \\ 0\le {\mathrm{R}}_{10}\le {\int }_{\mathrm{x}=10}^{\infty }{e}^{-x}\mathrm{dx} .....\left(1\right) \\ {\int }_{\mathrm{x}=10}^{\infty }{e}^{-x}\mathrm{dx} = e^{-10} \\ \mathrm{Therefore}, \\ 0\le {\mathrm{R}}_{10}\le e^{-10} \\ 0\le {\mathrm{R}}_{10}\le 0.000045 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{bound} \mathrm{on} {10}^{\mathrm{th}} \mathrm{remainder}, {\mathrm{R}}_{10} \mathrm{is} 0.000045 \end{gathered}$$

Part (d): 

$$\begin{gathered} {\mathrm{S}}_{10} \le \mathrm{L}\le {\mathrm{S}}_{10}+{\mathrm{B}}_{10} \\ 1.589150\le \mathrm{L}\le 1.589150+0.000045 \left(\mathrm{Substitution}\right) \\ 1.589150\le \mathrm{L}\le 1.581995 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{interval} \mathrm{containing} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{series} \mathrm{is} \mathrm{L} \in (1.589150, 1.581995). \end{gathered}$$

Part (e):

$$\begin{gathered} {\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} \le {10}^{-6} \\ {\mathrm{e}}^{-\mathrm{n}}\le {10}^{-6} \\ -\ln \left(\mathrm{e}\right) \le \ln \left({10}^{-6}\right) \\ \mathrm{n}\ge 13.8155 \\ \mathrm{n}\ge 14 (\mathrm{Round} \mathrm{to} \mathrm{nearest} \mathrm{integer}) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{n} \mathrm{so} \mathrm{that} {\mathrm{R}}_{\mathrm{n}}\le {10}^{-6} \mathrm{holds} \mathrm{is} \mathrm{n}=14 \end{gathered}$$

Step 1. Given information is:

$\sum _{k=0}^{\infty }{e}^{-k}$

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

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{e}}^{-\mathrm{x}}. \\ \mathrm{The} \mathrm{first} \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{is}: \\ \mathrm{f}\text{'}\left(\mathrm{x}\right) = -{\mathrm{e}}^{-\mathrm{x}} \\ \mathrm{The} \mathrm{derivative} \mathrm{is} \mathrm{negative} \mathrm{for} \mathrm{all} \mathrm{x}>1. \mathrm{Therefore}, \mathrm{the} \mathrm{function} \mathrm{is} \mathrm{decreasing}. \\ \mathrm{The} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{e}}^{-\mathrm{x}} \mathrm{is} \mathrm{continuous}, \mathrm{decreasing}, \mathrm{with} \mathrm{positive} \mathrm{terms}. \\ \mathrm{Therefore} \mathrm{all} \mathrm{the} \mathrm{conditions} \mathrm{of} \mathrm{integral} \mathrm{test} \mathrm{are} \mathrm{fulfilled}. \\ \mathrm{So}, \mathrm{integral} \mathrm{test} \mathrm{is} \mathrm{applicable}. \end{gathered}$$

Part (b) Step 2. Solving the integral:

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

Part (a) Step 3. Result

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{integral} \mathrm{is} {\int }_{\mathrm{x}=0}^{\infty }{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}=1 \\ \mathrm{The} \mathrm{integral} \mathrm{converges}. \\ \mathrm{Thus}, \mathrm{the} \mathrm{series} \underset{\mathrm{k}=0}{\overset{\infty }{\sum {\mathrm{e}}^{-k}}} \mathrm{is} \mathrm{convergent}. \end{gathered}$$

Part (b) Step 1. Finding S 10

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} {\mathrm{S}}_{10} \mathrm{is}: \\ {\mathrm{S}}_{10} = \frac{1}{e^0}+\frac{1}{e^1} +\frac{1}{e^2}+...+\frac{1}{e^9} \\ = \frac{1\left(1 - {\left({\displaystyle \frac{1}{e}}\right)}^{10}\right)}{1-{\displaystyle \frac{1}{e}}} \approx 1.589150 \\ \mathrm{Therefore}, \mathrm{the} {10}^{\mathrm{th}} \mathrm{term} \mathrm{in} \mathrm{the} \mathrm{sequence} \mathrm{of} \mathrm{partial} \mathrm{sums} \mathrm{to} \mathrm{approximate} \\ \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{series} \mathrm{is} {\mathrm{S}}_{10} \approx 1.589150 \end{gathered}$$

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

$$\begin{gathered} \mathrm{If} \mathrm{a} \mathrm{function} \mathrm{a} \mathrm{is} \mathrm{continuous}, \mathrm{positive}, \mathrm{and} \mathrm{decreasing}, \mathrm{and} \mathrm{if} \mathrm{the} \mathrm{improper} \\ \mathrm{Integral} {\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right) \mathrm{converges}, \mathrm{then} \mathrm{the} \mathrm{nth} \mathrm{remainder}, {\mathrm{R}}_{\mathrm{n}}, \mathrm{for} \mathrm{the} \mathrm{series} \\ \sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right) \mathrm{is} \mathrm{bounded} \mathrm{by}: \\ 0\le {\mathrm{R}}_{\mathrm{n}}\le {\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} \\ \mathrm{Furthermore}, \mathrm{if} {\mathrm{B}}_{\mathrm{n}} ={\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx}, \mathrm{then} \\ {\mathrm{S}}_{\mathrm{n}}\le \sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right) \le {\mathrm{S}}_{\mathrm{n}}+{\mathrm{B}}_{\mathrm{n}} \\ \mathrm{The} \mathrm{result} \mathrm{gives} \mathrm{that} \mathrm{the} \mathrm{tenth} \mathrm{remainder} {\mathrm{R}}_{10}, \mathrm{is} \mathrm{bounded} \mathrm{by}: \\ 0\le {\mathrm{R}}_{10}\le {\int }_{\mathrm{x}=10}^{\infty }{e}^{-x}\mathrm{dx} .....\left(1\right) \\ \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{integral} {\int }_{\mathrm{x}=10}^{\infty }{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx} \mathrm{is}: \\ {\int }_{\mathrm{x}=10}^{\infty }{e}^{-x}\mathrm{dx} = e^{-10} (\mathrm{Value} \mathrm{of} \mathrm{integral} \mathrm{solved} \mathrm{above} \mathrm{in} (\mathrm{a}\left)\right) \\ \mathrm{Therefore}, \mathrm{equation} \left(1\right) \mathrm{becomes}: \\ 0\le {\mathrm{R}}_{10}\le e^{-10} \\ 0\le {\mathrm{R}}_{10}\le 0.000045 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{bound} \mathrm{on} {10}^{\mathrm{th}} \mathrm{remainder}, {\mathrm{R}}_{10} \mathrm{is} 0.000045 \end{gathered}$$

Part (d) Step 1. Finding Interval

$$\begin{gathered} \mathrm{Assume} {\mathrm{B}}_{10}\approx 0.000045 \mathrm{and} \sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right)=\mathrm{L}. \\ \mathrm{Using} \mathrm{the} \mathrm{result} {\mathrm{S}}_{\mathrm{n}} \le \sum _{\mathrm{k}=1}^{\infty }\mathrm{a}\left(\mathrm{k}\right) \le {\mathrm{S}}_{\mathrm{n}} + {\mathrm{B}}_{\mathrm{n}}, \mathrm{the} \mathrm{following} \mathrm{inequality} \mathrm{is} \mathrm{obtained}. \\ {\mathrm{S}}_{10} \le \mathrm{L}\le {\mathrm{S}}_{10}+{\mathrm{B}}_{10} \\ 1.589150\le \mathrm{L}\le 1.589150+0.000045 \left(\mathrm{Substitution}\right) \\ 1.589150\le \mathrm{L}\le 1.581995 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{interval} \mathrm{containing} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{series} \mathrm{is} \mathrm{L} \in (1.589150, 1.581995). \end{gathered}$$

Part (e) Step 1. Finding n

$$\begin{gathered} \mathrm{To} \mathrm{find} \mathrm{the} \mathrm{smallest} \mathrm{value} \mathrm{of} \mathrm{n}, \mathrm{find} \mathrm{n} \mathrm{such} \mathrm{that} \mathrm{following} \mathrm{holds}: \\ {\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx}\le {10}^{-6} \\ {\mathrm{e}}^{-\mathrm{n}}\le {10}^{-6} \\ -\ln \left(\mathrm{e}\right) \le \ln \left({10}^{-6}\right) \\ \mathrm{n}\ge 13.8155 \\ \mathrm{n}\ge 14 (\mathrm{Round} \mathrm{to} \mathrm{nearest} \mathrm{integer}) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{n} \mathrm{so} \mathrm{that} {\mathrm{R}}_{\mathrm{n}}\le {10}^{-6} \mathrm{holds} \mathrm{is} \mathrm{n}=14 \end{gathered}$$