271s5

# 271s5 - MAT 271 Test #5 Solutions (1) (20 points) Find the...

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Unformatted text preview: MAT 271 Test #5 Solutions (1) (20 points) Find the radius of convergence for the series ∞ X k =5 ( x- 2) k k · 2 k . You do not need to find the interval of convergence. Solution: To find the radius of convergence for a power series, you use the ratio test (or, in some cases, you can use the root test). If a k = ( x- 2) k k · 2 k , then | a k +1 | | a k | = | x- 2 | k +1 ( k + 1) · 2 k +1 · k · 2 k | x- 2 | k = | x- 2 | 2 · k k + 1 , which converges to | x- 2 | 2 as k approaches infinity. The power series will converge if the limit of the ratio of consecutive terms is less than 1 and diverges if the limit is greater than 1 . Thus, if | x- 2 | < 2 , then the power series will converge for that value of x ; if | x- 2 | > 2 , it will diverge. Hence the radius of convergence is 2 (the 2 from the right-hand side of the equalities). Grading: +5 points for using the ratio test; +5 points for setting up the ratio; +5 points for setting the limit of the ratio less than 1 ; +5 points for getting the radius of convergence. (2) (15 points) Consider the series ∞ X n =3 4 n + 4 n 2 ( n + 2) 2 . Find the sum of this series, using the fact that 4 n + 4 n 2 ( n + 2) 2 = 1 n 2- 1 ( n + 2) 2 . Solution: The hint is that this series is a telescoping series. If you add up b n = 4 n + 4 n 2 ( n + 2) 2 for all n between 3 and N , then you get N X n =3 b n = 1 3 2- 1 5 2 + 1 4 2- 1 6 2 + 1 5...
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## This note was uploaded on 02/24/2009 for the course MAT 26996 taught by Professor Hurlbert during the Spring '08 term at ASU.

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271s5 - MAT 271 Test #5 Solutions (1) (20 points) Find the...

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