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 = 2x 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 = 2t
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) 1x xn xn =  (1)n1 n n
n=1 =2
n=1 n=1 2n1 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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