course Mth 174 vЅyY}tIǵassignment #013
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21:19:36 query problem 9.4.4 (9.3.6 3d edition). Using a comparison test determine whether the series sum(1/(3^n+1),n,1,infinity) converges.
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RESPONSE --> I concluded that 1 / (3^n + 1) <= 1 / (3^n) because the larger the denominator, the lower the value. From n=1 to n=inf, the series converges because sum(1 / x^p), p, 1, infinity) converges when p >= 1. And for all n values from 1 to infinity, they are greater than or equal to 1.
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21:19:43 With what known series did you compare this series, and how did you show that the comparison was valid?
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21:19:56 Query 9.4.10 3d edition 9.3.12). What is the radius of convergence of the series 1 / (2 n) ! and how did you use the ratio test to establish your result?
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RESPONSE --> lim(n -> inf) abs[a(n+1)] / abs[a(n)] = L. Converges when L < 1. a(n) = 1 / (2n)! a(n+1) = 1 / (2n + 1)! [1 / (2n + 1)!] / [1 / (2n)!] = (2n)! / (2n + 1)! = (2n)(2n - 1)(2n - 2).... / (2n + 1)(2n)(2n - 1)(2n - 2).... I cancel out (2n)(2n - 1)(2n - 2)..... to get 1 / (2n + 1) limit as n approaches infinity 1 / (2n + 1) = 0 So the series converges.
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21:20:10 Query problem 9.4.40 (3d edition 9.3.18) (was 9.2.24) partial sums of 1-.1+.01-.001 ... o what does the series converge?
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RESPONSE --> I didn't really grasp this part of the section. I know that a series converges if 0 < a(n+1) < a(n) and lim (n -> inf) a(n) = 0. In this series I set a(n) = (-1)^n * 10^(-n) and a(n +1) = -1^(n + 1) * 10^(-n + 1) Limit as n approaches infinity = 0 so it converges.
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21:20:20 What are the first five partial sums of the series?
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RESPONSE --> The first five partial sums are 1, .9, .91, .9109, .91091
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21:20:35 Query 9.5.6. What is your expression for the general term of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2) / 3! * x^3 + ?
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RESPONSE --> Since each term, after the first term, has a denominator I figured it to be (1 / n!). Each term is also increasing by x^n. The other part I was unsure of, but my guess was (p - n)! So my general term is: [p * (p - n)! * x^n] / (n + 1)!]
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21:20:45 Query 9.5.18. What is the radius of convergence of the series x / 3 + 2 x^2 / 5 + 3 x^2 / 7 + ?
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RESPONSE --> My general ratio for this is [n * x^n] / [1 + 2n] As n -> infinity, the limit of abs[a(n) + 1] / abs[ a(n)] = zero so raius equals infinity.
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21:20:50 What is your expression for the general term of this series, and how did you use this expression to find the radius of convergence?
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21:54:31 Query 9.5.28 (3d edition 9.4.24). What is the radius of convergence of the series p x + p(p-1) / 2! * x^2 + p(p-1)(p-2) * x^3 + and how did you obtain your result?
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RESPONSE --> I found the general term to be (p * p! * x^n) / n! To find the radius I put it in terms of abs(a(n) + 1) / abs(a(n)) a(n) = (p * p! * x^n) / n! a(n + 1) = (p * p! * x^(n + 1) / (n + 1)! abs(a(n) + 1) / abs(a(n)) = abs[p * p! * x^(n + 1) * n!] / abs[p * p! * x^n * (n + 1)!] limit as n approaches infinty is abs[p * p! * x] / abs[p * p! * (1)! I know the rules for finding a radius, but I am stuck here and do not know where to go.
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