course Mth 174 s݂k掼eassignment #013
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11:33:26 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 --> Looking at this sum, I can compare it to the sum of (1/3)^n. Looking at the limit of (1/3)^n indicates that as the value of n grows, the value of the sum grows toward 1/2 and seems to stay there.
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11:38:56 With what known series did you compare this series, and how did you show that the comparison was valid?
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RESPONSE --> I compared it with the series (1/3)^n. To make sure the comparison was valid, I looked at the limit of (1/3)^n and then made sure that 1/(3^n+1) was less than (1/3)^n.
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12:04:02 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 --> First I used the ratio test to see if the limit of this function was less or more than 1. lim( ( 1/(2(n+1))! ) / ( 1/(2n)! ), n->infinity) = lim( 1/(2(n+1)) * (2n)! / 1 ) = lim( (2n)! / ( (2(n+1))(2n)! ) ) = lim( 1 / (2(n + 1)) ) This can be split up into two limits: lim( 1/2 ) * lim( 1 / (n+1) ) The limit of 1 / (n+1) is equal to 0, and the limit of 1/2 is 1/2. Multiplied together, the overall limit is 0. Therefore this interval converges. Theorem 9.10 tells us that if the ratio between a_n+1 and a_n is 0, then the radius of convergence is infinite.
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12:21:17 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 --> The wild guess answer is 1 because the sums are oscillating above and below 1 in smaller and smaller increments. But to prove it, first I need to prove that it converges. To do this I need to make sure that each consecutive sum is less than the previous one: 1 > .1 > .01 > .001 So this seems to be true. Next I need to make sure that the limit of the formula is equal to 0: lim( (-1)^n * (10)^-n ) This goes to 0 as the value of n goes towards infinity. So this function does converge. Next the question is the value on which it converges. Using a calculator (and sum estimation), my estimate is the this sum converges on 1 -1/11 = .909.
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12:26:39 What are the first five partial sums of the series?
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RESPONSE --> S_1 = 1 - (1/10) = 0.9 S_2 = 1 - (9/100) = .91 S_3 = 1 - (91/1000) = .909 S_4 = 1 - (909/10000) = .9091 S_5 = 1 - (.09091) = .090909
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12:41:49 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) * x^3 + ?
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RESPONSE --> The only thing I could come up with is this: (p - (n-1)) x^n / n!
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13:35:06 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 --> The first thing I found was the formula: ( n x^n ) / (2n +1) Next I used the ratio test to find the ratio of the function: lim( ( ( (n+1) x^(n+1) ) / (2(n+1) +1) ) / ( ( n x^n ) / (2n +1) ) = lim( ( (n+1) x^(n+1) ) / (2(n+1) +1) * (2n+1) / (n x^n) = lim( (n+1)(2n+1)x / (2n+3)n ) = lim( (2n^2 +3n +1)x / 2n^2 +3n ) = | x | * lim( (2n^2 +3n +1) / 2n^2 ) For large values of n, the two 2n^2 terms will dominate, leaving a value like 2n^2 / 2n^2 = 1. Since the limit is 1, then K must also be equal to 1. Since a = 0, we can ignore it. K = 1 R = 1/K R = 1
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13:35:46 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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RESPONSE --> My expression was: n x^n / (2n+1) I think I explained how I used it above.
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14:11:48 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 --> Using the term I found before: (p - (n-1) ) x^n / n! Here is my result by using the ratio test: lim( ( (p - ((n+1)-1) ) x^(n+1) / (n+1)! ) / ( (p - (n-1) ) x^n / n!) = lim( (p - ((n+1)-1) ) x^(n+1) / (n+1)! ) * ( n! / (p - (n-1) ) ) = lim( (p - n ) x^n x / ( (n+1) n! ) * ( n! / (p - (n-1) ) ) = lim( (p-n)x / ( (p-(n+1))(n+1) ) = x * lim( (p - n) / ( -n^2 + pn + p) ) Since n / n^2 dominates, the function acts like 1/n for large values of n. But it is less than 1/n, so I can't compare it with 1/n to prove that it is divergent. Did I do something wrong with my original formula?
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