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 $\int_{1}^{\infty} a (x) d x$ converges, then the nth remainder, $R_{n}$ , for the series $\sum_{k = 1}^{\infty} a (k)$ is bounded by $0 \leq R_{n} = \sum_{k = n + 1}^{\infty} a (k) \leq \int_{n}^{\infty} a (x) d x$

Short Answer

Expert verified

$The function a : [1, \infty) \to ℝ is continuous and decreasing . Therefore : a (k) \leq \int_{x = k - 1}^{k} a (x) dx . . . . . (1) Hence, \sum_{k = n + 1}^{\infty} a (k) \leq \int_{x = n}^{\infty} a (x) dx (Summing (1)) . . . . . (2) From (2), 0 \leq R_{n} = \sum_{k = n + 1}^{\infty} a (k) \leq \int_{n}^{\infty} a (x) dx$

Step by step solution

Step 1. Given information is:

$\int_{1}^{\infty} a (x) dx is convergent$

Step 2. Finding Rn

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

Step 3. Result

$Also, the function a : [1, \infty) \to ℝ is continuous and decreasing . Therefore, following inequality holds : a (k) \leq \int_{x = k - 1}^{k} a (x) dx . . . . . (1) Hence, \sum_{k = n + 1}^{\infty} a (k) \leq \int_{x = n}^{\infty} a (x) dx (Summing (1)) . . . . . (2) From (2), it is proved that the n^{th} remainder, R_{n} for the series \sum_{k = 1}^{\infty} a_{k} is bounded by 0 \leq R_{n} = \sum_{k = n + 1}^{\infty} a (k) \leq \int_{n}^{\infty} a (x) d x$

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

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.