Q. 55 Prove Theorem 7.31. That is, sho... [FREE SOLUTION]

Chapter 7: Q. 55 (page 626)

Prove Theorem 7.31. That is, show that if a function a is continuous, positive, and decreasing, and if the improper integral ${鈭�}_{1}^{鈭�} a (x) d x$ converges, then the nth remainder, $R_{n}$ , for the series ${鈭�}_{k = 1}^{鈭�} a (k)$ is bounded by $0 鈮� R_{n} = {鈭�}_{k = n + 1}^{鈭�} a (k) 鈮� {鈭�}_{n}^{鈭�} a (x) d x$

Short Answer

Expert verified

$The function a : [1, 鈭�) 鈫� 鈩� is continuous and decreasing . Therefore : a (k) 鈮� {鈭�}_{x = k - 1}^{k} a (x) dx . . . . . (1) Hence, {鈭�}_{k = n + 1}^{鈭�} a (k) 鈮� {鈭�}_{x = n}^{鈭�} a (x) dx (Summing (1)) . . . . . (2) From (2), 0 鈮� R_{n} = {鈭�}_{k = n + 1}^{鈭�} a (k) 鈮� {鈭�}_{n}^{鈭�} a (x) dx$

Step by step solution

Step 1. Given information is:

${鈭�}_{1}^{鈭�} a (x) dx is convergent$

Step 2. Finding Rn

$The improper integral is convergent . Therefore, the series {鈭�}_{k = 1}^{鈭�} a_{k} is convergent by integral test and assume that it converges to sum L . The approximation of L with n^{th} partial sum S_{n} = {鈭�}_{k = 1}^{n} a_{k} gives the remainder R_{n} of the series {鈭�}_{k = 1}^{鈭�} a_{k} . The n^{th} remainder is defined by R_{n} and is given by : R_{n} = L - S_{n} = {鈭�}_{k = n + 1}^{鈭�} a_{k} Thus, the remainder R_{n} of the series {鈭�}_{k = 1}^{鈭�} a_{k} is R_{n} = {鈭�}_{k = n + 1}^{鈭�} a_{k} .$

Step 3. Result

$Also, the function a : [1, 鈭�) 鈫� 鈩� is continuous and decreasing . Therefore, following inequality holds : a (k) 鈮� {鈭�}_{x = k - 1}^{k} a (x) dx . . . . . (1) Hence, {鈭�}_{k = n + 1}^{鈭�} a (k) 鈮� {鈭�}_{x = n}^{鈭�} a (x) dx (Summing (1)) . . . . . (2) From (2), it is proved that the n^{th} remainder, R_{n} for the series {鈭�}_{k = 1}^{鈭�} a_{k} is bounded by 0 鈮� R_{n} = {鈭�}_{k = n + 1}^{鈭�} a (k) 鈮� {鈭�}_{n}^{鈭�} a (x) d x$

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Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Finding Rn

Step 3. Result

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91影视

Prove Theorem 7.31. That is, show that if a function a is continuous, positive, and decreasing, and if the improper integral 鈭�1鈭�a(x)dxconverges, then the nth remainder, Rn, for the series鈭�k=1鈭�a(k) is bounded by0鈮�Rn=鈭�k=n+1鈭�a(k)鈮�鈭�n鈭�a(x)dx

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Finding Rn

Step 3. Result

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Theoretical and Mathematical Physics

Statistics

Decision Maths

Pure Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.