Q. 54 - Calculus

Question

Let $a : [1,\infty )\to \mathrm{ℝ}$ be a continuous, positive, and decreasing function. Complete the proof of the integral test (Theorem 7.28) by showing that if the improper integral ${\int }_1^{\infty }a\left(x\right) dx$ converges, then the series $\sum _{\mathrm{k}=1 }^{\infty }a\left(k\right)$does too.

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{sequence} \mathrm{of} \mathrm{partial} \mathrm{sums} {\mathrm{s}}_{\mathrm{n}} = \sum _{\mathrm{i}=1}^{\infty } {\mathrm{a}}_{\mathrm{i}} \mathrm{is} \mathrm{bounded} \mathrm{above} \mathrm{by} \mathrm{M}. \\ \mathrm{The} \mathrm{terms} \mathrm{are} \mathrm{positive}. \mathrm{Therefore}, \\ {\mathrm{s}}_{\mathrm{n}} \le {\mathrm{s}}_{\mathrm{n}} + {\mathrm{a}}_{\mathrm{n}+1} \\ {\mathrm{s}}_{\mathrm{n}} \le {\mathrm{s}}_{\mathrm{n}+1} \\ \mathrm{The} \mathrm{sequence} \mathrm{of} \mathrm{partial} \mathrm{sum} \mathrm{is} \mathrm{an} \mathrm{increasing} \mathrm{sequence} \mathrm{and} \mathrm{is} \mathrm{bounded} \mathrm{above} \mathrm{by} \mathrm{M}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{sequence} \left\{{\mathrm{s}}_{\mathrm{n}}\right\} \mathrm{is} \mathrm{convergent}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } \mathrm{a}\left(\mathrm{k}\right) \mathrm{is} \mathrm{convergent}. \end{gathered}$$

Step 1. Given information is:

${\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right) \mathrm{dx} \mathrm{converges}$ and

$\mathrm{a} : [1,\infty )\to \mathrm{ℝ} \mathrm{is} \mathrm{continuous}, \mathrm{decreasing} \mathrm{and} \mathrm{positive} \mathrm{function}.$

Step 2. Evaluating integral

$$\begin{gathered} \mathrm{The} \mathrm{integral} {\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right) \mathrm{dx} \mathrm{is} \mathrm{convergent}. \mathrm{If} \mathrm{the} \mathrm{integral} \mathrm{is} \mathrm{convergent}, \\ \mathrm{then} \mathrm{the} \mathrm{integral} {\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} \mathrm{must} \mathrm{have} \mathrm{a} \mathrm{finite} \mathrm{value}. \\ \mathrm{Therefore}, \\ {\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right) \mathrm{dx} = \mathrm{M} \end{gathered}$$

Step 3. Solving for partial sum

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{function} \mathrm{a}\left(\mathrm{x}\right) \mathrm{is} \mathrm{positive}. \mathrm{Therefore}, \\ {\int }_1^{\mathrm{n}}\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} < {\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} .....\left(1\right) \\ \mathrm{Also}, \\ \sum _{\mathrm{i}=2}^{\infty } {\mathrm{a}}_{\mathrm{i}} < {\int }_1^{\mathrm{n}}\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} .....\left(2\right) \\ \mathrm{From} \mathrm{equations} \left(1\right) \mathrm{and} \left(2\right), \\ \sum _{\mathrm{i}=2}^{\infty } {\mathrm{a}}_{\mathrm{i}} < {\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} .....\left(3\right) \\ \mathrm{Also}, \\ \sum _{\mathrm{i}=1}^{\infty } {\mathrm{a}}_{\mathrm{i}} = {\mathrm{a}}_{1 }+ \sum _{\mathrm{i}=2}^{\infty } {\mathrm{a}}_{\mathrm{i}} \\ \mathrm{Using} \mathrm{equation} \left(3\right), \\ \sum _{\mathrm{i}=1}^{\infty } {\mathrm{a}}_{\mathrm{i}} < {\int }_1^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} \\ \sum _{\mathrm{i}=1}^{\infty } {\mathrm{a}}_{\mathrm{i}} <\mathrm{M} \end{gathered}$$

Step 4. Result

$$\begin{gathered} \mathrm{The} \mathrm{sequence} \mathrm{of} \mathrm{partial} \mathrm{sums} {\mathrm{s}}_{\mathrm{n}} = \sum _{\mathrm{i}=1}^{\infty } {\mathrm{a}}_{\mathrm{i}} \mathrm{is} \mathrm{bounded} \mathrm{above} \mathrm{by} \mathrm{M}. \\ \mathrm{The} \mathrm{terms} \mathrm{are} \mathrm{positive}. \mathrm{Therefore}, \\ {\mathrm{s}}_{\mathrm{n}} \le {\mathrm{s}}_{\mathrm{n}} + {\mathrm{a}}_{\mathrm{n}+1} \\ {\mathrm{s}}_{\mathrm{n}} \le {\mathrm{s}}_{\mathrm{n}+1} \\ \mathrm{The} \mathrm{sequence} \mathrm{of} \mathrm{partial} \mathrm{sum} \mathrm{is} \mathrm{an} \mathrm{increasing} \mathrm{sequence} \mathrm{and} \mathrm{is} \mathrm{bounded} \mathrm{above} \mathrm{by} \mathrm{M}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{sequence} \left\{{\mathrm{s}}_{\mathrm{n}}\right\} \mathrm{is} \mathrm{convergent}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } \mathrm{a}\left(\mathrm{k}\right) \mathrm{is} \mathrm{convergent}. \end{gathered}$$