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We are asked to find the values of p such that the following series converges. 4 2 n(In(n))P n = 2 We will use the Integral Test to test convergence. Recall that the Integral Test says that for a, = f(n), if f(x) is continuous, positive, and decreasing, then f(x) dx converges if and only if an converges. n = 2 We consider two cases: when p = 1 and when p + 1. 4 First assume that p = 1. We must check for f(n) = that f(x) = is continuous and positive on n In(n) the interval [2, ∞). The functions y = x and y = In(x) are continuous and positive on the interval, so f(x) is as x In(x) well. We can show that f'(x) 1 + In(x) = -4 such that 1 + In(x) > 0, which is the f'(x) < 0 for all x²(In(x))²´ case for the given interval x > 2, so f(x) is decreasing. Therefore, the Integral Test applies, and we evaluate the integral. 4 4 lim In dx = x In(x) 12 - In(in(2) = 4 lim In In(In(2)) = 00 4 As the integral diverges, for p = 1, the series ---Select--- n In(n) n = 2

We are asked to find the values of p such that the following series converges. 4 2 n(In(n))P n = 2 We will use the Integral Test to test convergence. Recall that the Integral Test says that for a, = f(n), if f(x) is continuous, positive, and decreasing, then f(x) dx converges if and only if an converges. n = 2 We consider two cases: when p = 1 and when p + 1. 4 First assume that p = 1. We must check for f(n) = that f(x) = is continuous and positive on n In(n) the interval [2, ∞). The functions y = x and y = In(x) are continuous and positive on the interval, so f(x) is as x In(x) well. We can show that f'(x) 1 + In(x) = -4 such that 1 + In(x) > 0, which is the f'(x) < 0 for all x²(In(x))²´ case for the given interval x > 2, so f(x) is decreasing. Therefore, the Integral Test applies, and we evaluate the integral. 4 4 lim In dx = x In(x) 12 - In(in(2) = 4 lim In In(In(2)) = 00 4 As the integral diverges, for p = 1, the series ---Select--- n In(n) n = 2

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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We are asked to find the values of p such that the following series converges. 00 4 Σ n(In(n))P n = 2 We will use the Integral Test to test convergence. Recall that the Integral Test says that for a, f(n), if f(x) is 00 continuous, positive, and decreasing, then f(x) dx converges if and only if a, converges. n = 2 We consider two cases: when p = 1 and when p # 1. = 1. We must check for f(n) 4 that f(x) 4 is continuous and positive on First assume that p %3D x In(x) n In(n) the interval [2, 0). The functions y = x and y = In(x) are continuous and positive on the interval, so f(x) is as 1 + In(x) well. We can show that f'(x) = -4 Then f'(x) < 0 for all x such that 1 + In(x) > 0, which is the x²(In(x))² case for the given interval x > 2, so f(x) is decreasing. Therefore, the Integral Test applies, and we evaluate the integral. 4 dx 4 lim %D x In(x) 12 = 4 lim In In (In(2)) = 00 4 As the integral diverges, for p = 1, the series ---Select--- n In(n) n = 2
Transcribed Image Text:We are asked to find the values of p such that the following series converges. 00 4 Σ n(In(n))P n = 2 We will use the Integral Test to test convergence. Recall that the Integral Test says that for a, f(n), if f(x) is 00 continuous, positive, and decreasing, then f(x) dx converges if and only if a, converges. n = 2 We consider two cases: when p = 1 and when p # 1. = 1. We must check for f(n) 4 that f(x) 4 is continuous and positive on First assume that p %3D x In(x) n In(n) the interval [2, 0). The functions y = x and y = In(x) are continuous and positive on the interval, so f(x) is as 1 + In(x) well. We can show that f'(x) = -4 Then f'(x) < 0 for all x such that 1 + In(x) > 0, which is the x²(In(x))² case for the given interval x > 2, so f(x) is decreasing. Therefore, the Integral Test applies, and we evaluate the integral. 4 dx 4 lim %D x In(x) 12 = 4 lim In In (In(2)) = 00 4 As the integral diverges, for p = 1, the series ---Select--- n In(n) n = 2
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