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Question 3 end. Read the short write-up on the integral test and answer the question at the First, we use an example to demonstrate the idea of the integral test. Consider the series Σ 72=1 1 n2 We can geometrically represent each term 1/n² as the area of a rectangle with width 1 and height 1/n². In Figure 3, the sequence {1/n²} is drawn as blue dots, and the rectangles corresponding to 1 1 FIGURE 3. The sequence n2 5 } and the function f(x) the first 5 terms are drawn in light blue. Therefore, the sum of the series of the area of rectangles representing the terms of the series. 1 n² is the infinite sum Notice the function f(x) satisfies two conditions: x2 (a) For every integer n ≥ 1, we have f(n) ========= which is the n-th term of the series. (b) The improper integral © ƒ(x) dx is convergent. The first condition implies that the region between y = f(x) and y = 0 for x ≥ 1 completely contains all rectangles (expect the first one) representing the terms of the series. It follows that f(x) dx = Σ 1 n2- n2 must also be convergent. Since the improper integral ƒ(x) dx is convergent, then the series The Integral Test describes the phenomenon in a general setting. Theorem (The Integral Test). Suppose f is a continuous, positive, decreasing function on domain [1, ∞] and let an = f(n) for every positive integer n. Then (i) If √ f(x) dx is convergent, then Σ is convergent. f(x) dx is divergent, then Σ is divergent. (ii) If n=1 In(n) is convergent. Use the integral test to determine all values of p for which the series nP All solutions need to use the integral test 3

Question 3 end. Read the short write-up on the integral test and answer the question at the First, we use an example to demonstrate the idea of the integral test. Consider the series Σ 72=1 1 n2 We can geometrically represent each term 1/n² as the area of a rectangle with width 1 and height 1/n². In Figure 3, the sequence {1/n²} is drawn as blue dots, and the rectangles corresponding to 1 1 FIGURE 3. The sequence n2 5 } and the function f(x) the first 5 terms are drawn in light blue. Therefore, the sum of the series of the area of rectangles representing the terms of the series. 1 n² is the infinite sum Notice the function f(x) satisfies two conditions: x2 (a) For every integer n ≥ 1, we have f(n) ========= which is the n-th term of the series. (b) The improper integral © ƒ(x) dx is convergent. The first condition implies that the region between y = f(x) and y = 0 for x ≥ 1 completely contains all rectangles (expect the first one) representing the terms of the series. It follows that f(x) dx = Σ 1 n2- n2 must also be convergent. Since the improper integral ƒ(x) dx is convergent, then the series The Integral Test describes the phenomenon in a general setting. Theorem (The Integral Test). Suppose f is a continuous, positive, decreasing function on domain [1, ∞] and let an = f(n) for every positive integer n. Then (i) If √ f(x) dx is convergent, then Σ is convergent. f(x) dx is divergent, then Σ is divergent. (ii) If n=1 In(n) is convergent. Use the integral test to determine all values of p for which the series nP All solutions need to use the integral test 3

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 22E
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hello can someone please help me with this calculus question, thank you!

Question 3 end. Read the short write-up on the integral test and answer the question at the First, we use an example to demonstrate the idea of the integral test. Consider the series Σ 72=1 1 n2 We can geometrically represent each term 1/n² as the area of a rectangle with width 1 and height 1/n². In Figure 3, the sequence {1/n²} is drawn as blue dots, and the rectangles corresponding to 1 1 FIGURE 3. The sequence n2 5 } and the function f(x) the first 5 terms are drawn in light blue. Therefore, the sum of the series of the area of rectangles representing the terms of the series. 1 n² is the infinite sum Notice the function f(x) satisfies two conditions: x2 (a) For every integer n ≥ 1, we have f(n) ========= which is the n-th term of the series. (b) The improper integral © ƒ(x) dx is convergent. The first condition implies that the region between y = f(x) and y = 0 for x ≥ 1 completely contains all rectangles (expect the first one) representing the terms of the series. It follows that f(x) dx = Σ 1 n2- n2 must also be convergent. Since the improper integral ƒ(x) dx is convergent, then the series The Integral Test describes the phenomenon in a general setting. Theorem (The Integral Test). Suppose f is a continuous, positive, decreasing function on domain [1, ∞] and let an = f(n) for every positive integer n. Then (i) If √ f(x) dx is convergent, then Σ is convergent. f(x) dx is divergent, then Σ is divergent. (ii) If n=1 In(n) is convergent. Use the integral test to determine all values of p for which the series nP All solutions need to use the integral test 3
Transcribed Image Text:Question 3 end. Read the short write-up on the integral test and answer the question at the First, we use an example to demonstrate the idea of the integral test. Consider the series Σ 72=1 1 n2 We can geometrically represent each term 1/n² as the area of a rectangle with width 1 and height 1/n². In Figure 3, the sequence {1/n²} is drawn as blue dots, and the rectangles corresponding to 1 1 FIGURE 3. The sequence n2 5 } and the function f(x) the first 5 terms are drawn in light blue. Therefore, the sum of the series of the area of rectangles representing the terms of the series. 1 n² is the infinite sum Notice the function f(x) satisfies two conditions: x2 (a) For every integer n ≥ 1, we have f(n) ========= which is the n-th term of the series. (b) The improper integral © ƒ(x) dx is convergent. The first condition implies that the region between y = f(x) and y = 0 for x ≥ 1 completely contains all rectangles (expect the first one) representing the terms of the series. It follows that f(x) dx = Σ 1 n2- n2 must also be convergent. Since the improper integral ƒ(x) dx is convergent, then the series The Integral Test describes the phenomenon in a general setting. Theorem (The Integral Test). Suppose f is a continuous, positive, decreasing function on domain [1, ∞] and let an = f(n) for every positive integer n. Then (i) If √ f(x) dx is convergent, then Σ is convergent. f(x) dx is divergent, then Σ is divergent. (ii) If n=1 In(n) is convergent. Use the integral test to determine all values of p for which the series nP All solutions need to use the integral test 3
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