sample_midterm2_21c_solutions

# sample_midterm2_21c_solutions - Calculus Math 21C Fall 2010...

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Unformatted text preview: Calculus : Math 21C, Fall 2010 Solutions to Sample Questions: Midterm 2 1. Determine the interval of convergence (including the endpoints) for the following power series. State explicitly for what values of x the series con- verges absolutely, converges conditionally, and diverges. In each case, specify the radius of convergence R and the center of the interval of convergence a . (a) ∞ X n =1 (- 1) n 2 n n ( x- 1) n ; (b) ∞ X n =0 1 3 n + 1 x 2 n ; (c) ∞ X n =0 1 n 2 5 n (2 x + 1) n . Solution. • We give two different (but related) methods for finding the solution. Either one is fine. • Method I. Write the power series as ∑ ∞ n =0 a n . • (a) We have lim n →∞ a n +1 a n = lim n →∞ (- 1) n +1 2 n +1 ( x- 1) n +1 / ( n + 1) (- 1) n 2 n ( x- 1) n /n = lim n →∞ 2( x- 1) n n + 1 = 2 | x- 1 | lim n →∞ n n + 1 = 2 | x- 1 | lim n →∞ 1 1 + 1 /n = 2 | x- 1 | . By the ratio test, the series converges absolutely when 2 | x- 1 | < 1 or | x- 1 | < 1 / 2. • Therefore, a = 1 and R = 1 / 2. The series converges absolutely if 1 / 2 < x < 3 / 2 and diverges if-∞ < x < 1 / 2 or 3 / 2 < x < ∞ . 1 • At the endpoint x = 3 / 2 of the interval of convergence, the series becomes ∞ X n =1 (- 1) n n . This is the alternating harmonic series. It converges by the alternating series test but does not converge absolutely since the harmonic series diverges. Therefore, the series converges conditionally at x = 3 / 2. • At the endpoint x = 1 / 2 of the interval of convergence, the series becomes- ∞ X n =1 1 n , which a harmonic series. Therefore, the series diverges at x = 1 / 2. • (b) We have lim n →∞ a n +1 a n = lim n →∞ x 2( n +1) / (3 n +1 + 1) x 2 n / (3 n + 1) = lim n →∞ x 2 (3 n + 1) 3 n +1 + 1 = x 2 lim n →∞ 3 n + 1 3 n +1 + 1 = x 2 lim n →∞ 1 + 1 / 3 n 3 + 1 / 3 n = x 2 3 . By the ratio test, the series converges absolutely when x 2 3 < 1 or | x | < √ 3. • Therefore, a = 0 and R = √ 3. The series converges absolutely if | x | < √ 3 and diverges if | x | > √ 3. 2 • At the endpoints x = ± √ 3 of the interval of convergence, the series becomes ∞ X n =0 3 n 3 n + 1 . Since lim n →∞ 3 n 3 n + 1 = 1 , this series diverges by the n th term test. • (c) We have lim n →∞ a n +1 a n = lim n →∞ (2 x + 1) n +1 / [( n + 1) 2 5 n +1 ] (2 x + 1) n / [ n 2 5 n ] = lim n →∞ (2 x + 1) n 2 5( n + 1) 2 = | 2 x + 1 | 5 lim n →∞ n 2 ( n + 1) 2 = | 2 x + 1 | 5 lim n →∞ 1 (1 + 1 /n ) 2 = | 2 x + 1 | 5 . By the ratio test, the series converges absolutely when | 2 x + 1 | 5 < 1 or | x + 1 / 2 | < 5 / 2....
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sample_midterm2_21c_solutions - Calculus Math 21C Fall 2010...

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