fa07hw10

# fa07hw10 - 19 Apply the ratio test to-1)n(x 1)n 2n 3(x 1)n...

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Unformatted text preview: 19. Apply the ratio test to (-1)n (x - 1)n : 2n + 3 (x - 1)n+1 2n + 3 = |x - 1|. 2n + 5 (x - 1)n = lim The series converges absolutely if |x - 1| < 1, that is, if 0 < x < 2, and converges conditionally if x = 2. It diverges for all other values of x. 21. Apply the ratio test to xn : 2n ln n x n+1 2n ln n ln n |x| |x| lim = . = n+1 ln(n + 1) n 2 x 2 ln(n + 1) 2 = lim (The last limit can be evaluated by l'H^ pital's Rule.) The given series converges absolutely o if |x| < 2, that is, if -2 < x < 2. By the alternating series test, it converges conditionally if x = -2. It diverges for all other values of x. n+2 = 1. The radius of convergence is 1; the we have R = lim 1. For n+1 n+1 n=0 centre of convergence is 0; the interval of convergence is (-1, 1). (The series does not converge at x = -1 or x = 1.) x 2n 1 x +2 n 2n+1 (n + 1) 3. For = 2. The radius of convergence is 2; we have R = lim n 2 2n n n=1 the centre of convergence is -2. For x = -4 the series is an alternating harmonic series, so converges. For x = 0, the series is a divergent harmonic series. Therefore the interval of convergence is [-4, 0). 5. n=0 n (2x - 3) = n=0 3 n 2n n 3 x - 3 n 2 . Here 1 2n n 3 = . The radius of convergence is 1/2; the centre of convergence n+1 (n + 1)3 2 2 is 3/2; the interval of convergence is (1, 2). R = lim 13. 1 1 1 = 2-x 2 1- x 2 x x2 x3 1 = + 2 + 3 + 4 2 2 2 2 for -2 < x < 2. Now differentiate to get 1 1 2x 3x 2 = 2 + 3 + 4 + (2 - x)2 2 2 2 = n=0 (n + 1)x n , 2n+2 (-2 < x < 2). x 15. 0 dt = 2-t x x tn - ln(2 - t) 0 = n+1 0 n=0 2 t n+1 n=0 n=0 dt x 0 2n+1 (n + 1) x n+1 2n+1 (n + 1) - ln(2 - x) + ln 2 = ln(2 - x) = ln 2 - n=1 xn . (-2 x < 2). 2n n 1. e 3x+1 =ee 3x =e = n=0 e3n x n (3x)n n! (for all x). n=0 n! 3. sin x - 1 = 2 1 = 2 4 n=0 n=0 - cos x sin 4 4 2n+1 x x 2n 1 (-1)n (-1)n - (2n + 1)! (2n)! 2 = sin x cos (-1)n - x 2n (2n)! + n=0 2n+1 x (2n + 1)! (for all x). 7. sin x cos x = = 1 sin(2x) 2 (-1)n n=0 22n x 2n+1 (2n + 1)! (for all x). 11. ln 1+x = ln(1 + x) - ln(1 - x) 1-x xn xn = - (-1)n-1 n n n=1 =2 n=1 n=1 2n-1 x 2n - 1 (-1 < x < 1). 1. (-1)n converges by the alternating series test (since the terms alternate in sign, decrease n 1 in size, and approach 0). However, the convergence is only conditional, since n diverges to infinity. 5. (-1)n (n 2 - 1) diverges since its terms do not approach zero. n2 + 1 7. (-1)n converges absolutely, since, for n 1, n n (-1)n 1 n, n n and 1 is a convergent geometric series. n 17. Applying the ratio test to xn , we obtain n+1 n+1 = |x| lim xn n+1 = |x|. n+2 x n+1 = lim n+2 Hence the series converges absolutely if |x| < 1, that is, if -1 < x < 1. The series converges conditionally for x = -1, but diverges for all other values of x. ...
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• Fall '06
• GROSS
• Calculus, Mathematical Series, Convergence, Tn

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