Homework7Solns

# Homework7Solns - SHORT CALCULUS Math 16C Sec 2 Spring 2008...

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Unformatted text preview: SHORT CALCULUS Math 16C Sec 2 Spring 2008 Homework #7 Solutions Peter Malkin Section 10.4 Question 4 ∞ X n =1 (- 1) n x n ( n- 1)! =- 1 + x 2- x 3 2 + x 4 6- x 5 24 + ... Question 10 lim n →∞ 3 n +1 / ( n + 1)! 3 n /n ! · x n +1 x n = lim n →∞ 3 n +1 3 n · n ! ( n + 1)! · x = lim n →∞ 3 · 1 n + 1 x = 0 By the ratio test, this series converges for all x , so the radius of convergence is ∞ . Question 14 lim n →∞ (- 1) n +1 ( n + 1)! / 3 n +1 (- 1) n n ! / 3 n · ( x- 4) n +1 ( x- 4) n = lim n →∞ (- 1) n +1 (- 1) n · ( n + 1)! n ! · 3 n 3 n +1 · ( x- 4) = lim n →∞ (- 1) · ( n + 1) · 1 3 · ( x- 4) = ∞ ( x 6 = 4) By the ratio test, this series converges for all x 6 = 4, so the radius of convergence is 0. Question 16 lim n →∞ 1 / ( ( n + 2)3 n +2 ) 1 / (( n + 1)3 n +1 ) · ( x- 2) n +2 ( x- 2) n +1 = lim n →∞ n + 1 n + 2 · 3 n +1 3 n +2 · ( x- 2) = lim n →∞ 1 + 1 n 1 + 2 n · 1 3 · ( x- 2) = 1 3 ( x- 2) By the ratio test, this series converges for all x such that 1 3 ( x- 2) < 1 ⇔ | x- 2 | < 3, so the radius of convergence is 3....
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Homework7Solns - SHORT CALCULUS Math 16C Sec 2 Spring 2008...

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