course MTH 174 8:10 p.m. 7/11/10. û믹}zߺYassignment #013
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19:10:50 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 --> Here we could do a sum comparison using a(n+1)/an. But I used an integral comparison with: INT(1/3^N, 1, inf). The integral is -1/(ln 3 3^n) evaluating over the interval gives us: -1/(ln 3 3^b) - -1/(ln 3 3) as b approaches infinity the first term goes to zero leaving: 1/((ln 3(3)) which is approximately 0.30413. So the integral converges and we expect the series to follow suit. Setting up our inequality, we have: 1/(3^n+1)<= 1/3^n, then: 3^n <= 3^n+1 which is always true on our interval so the series does indeed converge.
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19:13:25 With what known series did you compare this series, and how did you show that the comparison was valid?
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RESPONSE --> The comparison series is 1/3^n but in the above solution I used an integral test with the same expression and we could also have used a ratio test.
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19:20:06 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 --> Using the ratio test with a(n+1)/a(n) where a(n+1)= 1/(2n+2)! and a(n)= 1/(2n)!. The ratio is then: (1/(2n+2)!)/(1/2n!). Some algebra makes this: 2n!/(2n+2)!. Expanding and dividing leaves 1/((2n+2)(2n+1)) The limit as n-> inf is zero and the ratio theorem with L<1 says that our series converges. Also, since the ratio turned out to be zero the Theorem 9.10 says that the radius is convergence is: R= inf.
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19:35:07 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 alternating series theorem tells that if 0< a(n+1)< a(n) and the limit of a(n) as n-> inf = 0 then the series converges. Here a(n)= 1/10^n and then a(n+1)= 1/10^(n+1) and 0< 1/10^(n+1) < 1/10^n and as n -> inf. the expression goes to zero. So the Alternating series series test says the series converges. The convergent value can be estimated by looking at the values the series sum takes on: 1, .9, 91, .909, .9091, .90909. We can see the pattern here and say that the convergent value will be about 0.909090909....
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19:36:47 What are the first five partial sums of the series?
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RESPONSE --> In parenthetical form the values are: (1,1)(2,0.9)(3,0.91)(4, 0.909)(5, 0.90909).
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19:41:48 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 --> It is easy to see x^n and 1/n! but the p, p(p-1), p(p-1)(p-2) parts are a little trickier. It clearly involves p! but we have to get rid of the extra terms in the expansion by division. (p-n)! looks good because that leaves p and then p(p-1) etc. as n increases. So the expression is (p!/(p-n)!) x^n.
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20:01:15 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 --> Our general expression is clear here and is: n/(2n+1) x^n. Using the ratio test with a(n)=n/(2n+1) and a(n+1)= (n+1)/(2n+3). The ratio is then [(n+1)/(2n+3)]/[n/(2n+1)]. A little rearranging makes this: [n/(2n+1)]* [(2n+3)/(n+1)] Here multiplication makes this: (2n^2+3n)/(2n^2+n+2n+1). Division reduces this to 1 and so the limit is also 1 and then the radius of convergence is 1/1 or 1.
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20:02:18 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 --> The expression is: (n/(2n+1)) x^n. As shown above I used the ratio test.
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20:11:29 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 --> With a(n)= (p!/p-n)!)/n! and then a(n+1)=(p!/(p-n-1)!)/(n+1)!. Some multiplication and reciprocation makes this: [p!/((n+1)!(p-n-1)!)]*[(n!(p-n)!)/p!]. The p! terms divide out and then expanding the ! terms and dividing leaves: (p-n)/(n+1) p-n as n-> inf goes to -inf and n+1 goes to +inf and dividing the two makes -1. The radius is then 1/1 or 1.
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