MT2Solutions - Solutions to Midterm 2 1. Consider the power...

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Solutions to Midterm 2 1. Consider the power series X n =1 (4 x ) n n . a) (10 points) Find the radius of convergence. b) (10 points) Find the interval of convergence. Answer: R = 1 / 4, interval [ - 1 4 , 1 4 ) . Solution: a) Let us apply the Ratio Test. We have a n = (4 x ) n n , a n +1 = (4 x ) n +1 n + 1 , so lim n →∞ ± ± ± ± a n +1 a n ± ± ± ± = lim n →∞ ± ± ± ± ± (4 x ) n +1 (4 x ) n ± ± ± ± ± · n n + 1 = | 4 x | . Therefore the series converges if | 4 x | < 1 ⇔ | x | < 1 4 , and diverges, if | x | > 1 4 , so the radius of convergence is equal to 1 4 . b) Let us check the endpoints. For x = 1 4 we get a series X n =1 (4 / 4) n n = X n =1 1 n . This is a divergent p -series with p = 1. For x = - 1 4 we get a series X n =1 ( - 4 / 4) n n = X n =1 ( - 1) n n . It converges by the Alternating Series Test, as 1 /n is a decreasing sequence converging to 0. Grading: -1 point, if the Ratio Test is correct, but the radius is incorrect; 2 points for correct series in each endpoint and 3 points for the convergence test in each endpoint 2. Consider the power series X n =1 ( - 1) n ( x/ 2) n n 2 . 1
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a) (10 points) Find the radius of convergence. b) (10 points) Find the interval of convergence. Answer: R = 2, interval [ - 2 , 2] . Solution: a) Let us apply the Ratio Test. We have a n = ( - 1) n ( x/ 2) n n 2 , a n +1 = ( - 1) n +1 ( x/ 2) n +1 ( n + 1) 2 , so lim n →∞ ± ± ± ± a n +1 a n ± ± ± ± = lim n →∞ ± ± ± ± ± ( x/ 2) n +1 ( x/ 2) n ± ± ± ± ± · n 2 ( n + 1) 2 = | x/ 2 | . Therefore the series converges if | x/ 2 | < 1 ⇔ | x | < 2, and diverges, if | x | > 2, so the radius of convergence is equal to 2. b) Let us check the endpoints. For x = 2 we get a series X n =1 ( - 2 / 2) n n 2 = X n =1 ( - 1) n n . It converges by the Alternating Series Test, as
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MT2Solutions - Solutions to Midterm 2 1. Consider the power...

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