Q. 55 - Calculus

Question

Prove Theorem 7.31. That is, show that if a function a is continuous, positive, and decreasing, and if the improper integral ${\int }_1^{\infty }a\left(x\right)dx$ converges, then the nth remainder, $R_n$, for the series $\sum _{k=1}^{\infty } a\left(k\right)$ is bounded by $0 \le R_{n }= \sum _{k=n+1}^{\infty } a\left(k\right) \le {\int }_n^{\infty }a\left(x\right)dx$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{function} \mathrm{a}:[1, \infty )\to \mathrm{ℝ} \mathrm{is} \mathrm{continuous} \mathrm{and} \mathrm{decreasing}. \\ \mathrm{Therefore}: \\ \mathrm{a}\left(\mathrm{k}\right) \le {\int }_{\mathrm{x}=\mathrm{k}-1}^{\mathrm{k}}\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} .....\left(1\right) \\ \mathrm{Hence}, \\ \sum _{\mathrm{k}=\mathrm{n}+1}^{\infty } \mathrm{a}\left(\mathrm{k}\right) \le {\int }_{\mathrm{x}=\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} ( \mathrm{Summing} (1\left)\right) .....\left(2\right) \\ \mathrm{From} \left(2\right), \\ 0 \le {\mathrm{R}}_{\mathrm{n} }= \sum _{\mathrm{k}=\mathrm{n}+1}^{\infty } \mathrm{a}\left(\mathrm{k}\right) \le {\int }_{\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} \end{gathered}$$

Step 1. Given information is:

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

Step 2. Finding R n

$$\begin{gathered} \mathrm{The} \mathrm{improper} \mathrm{integral} \mathrm{is} \mathrm{convergent}. \mathrm{Therefore}, \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{a}}_{\mathrm{k}} \mathrm{is} \mathrm{convergent} \\ \mathrm{by} \mathrm{integral} \mathrm{test} \mathrm{and} \mathrm{assume} \mathrm{that} \mathrm{it} \mathrm{converges} \mathrm{to} \mathrm{sum} \mathrm{L}. \\ \mathrm{The} \mathrm{approximation} \mathrm{of} \mathrm{L} \mathrm{with} {\mathrm{n}}^{\mathrm{th}} \mathrm{partial} \mathrm{sum} {\mathrm{S}}_{\mathrm{n}} = \sum _{\mathrm{k}=1}^{\mathrm{n}} {\mathrm{a}}_{\mathrm{k}} \mathrm{gives} \mathrm{the} \\ \mathrm{remainder} {\mathrm{R}}_{\mathrm{n}} \mathrm{of} \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{a}}_{\mathrm{k}} . \\ \mathrm{The} {\mathrm{n}}^{\mathrm{th}} \mathrm{remainder} \mathrm{is} \mathrm{defined} \mathrm{by} {\mathrm{R}}_{\mathrm{n} }\mathrm{and} \mathrm{is} \mathrm{given} \mathrm{by}: \\ {\mathrm{R}}_{\mathrm{n}} = \mathrm{L} - {\mathrm{S}}_{\mathrm{n}} \\ = \sum _{\mathrm{k}=\mathrm{n}+1}^{\infty } {\mathrm{a}}_{\mathrm{k}} \\ \mathrm{Thus}, \mathrm{the} \mathrm{remainder} {\mathrm{R}}_{\mathrm{n}} \mathrm{of} \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{a}}_{\mathrm{k}} \mathrm{is} {\mathrm{R}}_{\mathrm{n}} = \sum _{\mathrm{k}=\mathrm{n}+1}^{\infty } {\mathrm{a}}_{\mathrm{k}} . \end{gathered}$$

Step 3. Result

$$\begin{gathered} \mathrm{Also}, \mathrm{the} \mathrm{function} \mathrm{a}:[1, \infty )\to \mathrm{ℝ} \mathrm{is} \mathrm{continuous} \mathrm{and} \mathrm{decreasing}. \\ \mathrm{Therefore}, \mathrm{following} \mathrm{inequality} \mathrm{holds}: \\ \mathrm{a}\left(\mathrm{k}\right) \le {\int }_{\mathrm{x}=\mathrm{k}-1}^{\mathrm{k}}\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} .....\left(1\right) \\ \mathrm{Hence}, \\ \sum _{\mathrm{k}=\mathrm{n}+1}^{\infty } \mathrm{a}\left(\mathrm{k}\right) \le {\int }_{\mathrm{x}=\mathrm{n}}^{\infty }\mathrm{a}\left(\mathrm{x}\right)\mathrm{dx} ( \mathrm{Summing} (1\left)\right) .....\left(2\right) \\ \mathrm{From} \left(2\right), \mathrm{it} \mathrm{is} \mathrm{proved} \mathrm{that} \mathrm{the} {\mathrm{n}}^{\mathrm{th}} \mathrm{remainder}, {\mathrm{R}}_{\mathrm{n}} \mathrm{for} \mathrm{the} \mathrm{series} \sum _{\mathrm{k}=1}^{\infty } {\mathrm{a}}_{\mathrm{k} }\mathrm{is} \mathrm{bounded} \mathrm{by} \\ 0 \le R_{n }= \sum _{k=n+1}^{\infty } a\left(k\right) \le {\int }_n^{\infty }a\left(x\right)dx \end{gathered}$$