D from example 7 in section 129 et 119 we have arctan

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(d) From Example 7 in Section 12.9 [ ET 11.9] we have arctan x = x x 3 3 + x 5 5 x 7 7 + x 9 9 x 11 11 + · · · , so arctan 1 5 = 1 5 1 3 · 5 3 + 1 5 · 5 5 1 7 · 5 7 + 1 9 · 5 9 1 11 · 5 11 + · · · This is an alternating series and the size of the terms decreases to 0 , so by the Alternating Series Estimation Theorem, the sum lies between s 5 and s 6 , that is, 0 . 197395560 < arctan 1 5 < 0 . 197395562 . (e) From the series in part (d) we get arctan 1 239 = 1 239 1 3 · 239 3 + 1 5 · 239 5 · · · . The third term is less than 2 . 6 × 10 13 , so by the Alternating Series Estimation Theorem, we have, to nine decimal places, arctan 1 239 s 2 0 . 004184076 . Thus, 0 . 004184075 < arctan 1 239 < 0 . 004184077 .
PROBLEMS PLUS ¤ 101 (f) From part (c) we have π = 16 arctan 1 5 4 arctan 1 239 , so from parts (d) and (e) we have 16(0 . 197395560) 4(0 . 004184077) < π < 16(0 . 197395562) 4(0 . 004184075) 3 . 141592652 < π < 3 . 141592692 . So, to 7 decimal places, π 3 . 1415927 . 9. We start with the geometric series S n =0 x n = 1 1 x , | x | < 1 , and differentiate: S n =1 nx n 1 = d dx · S n =0 x n ¸ = d dx · 1 1 x ¸ = 1 (1 x ) 2 for | x | < 1 S n =1 nx n = x S n =1 nx n 1 = x (1 x ) 2 for | x | < 1 . Differentiate again: S n =1 n 2 x n 1 = d dx x (1 x ) 2 = (1 x ) 2 x · 2(1 x )( 1) (1 x ) 4 = x + 1 (1 x ) 3 S n =1 n 2 x n = x 2 + x (1 x ) 3 S n =1 n 3 x n 1 = d dx x 2 + x (1 x ) 3 = (1 x ) 3 (2 x + 1) ( x 2 + x )3(1 x ) 2 ( 1) (1 x ) 6 = x 2 + 4 x + 1 (1 x ) 4 S n =1 n 3 x n = x 3 + 4 x 2 + x (1 x ) 4 , | x | < 1 . The radius of convergence is 1 because that is the radius of convergence for the geometric series we started with. If x = ± 1 , the series is S n 3 ( ± 1) n , which diverges by the Test For Divergence, so the interval of convergence is ( 1 , 1) . 11. ln · 1 1 n 2 ¸ = ln · n 2 1 n 2 ¸ = ln ( n + 1)( n 1) n 2 = ln[( n + 1)( n 1)] ln n 2 = ln( n + 1) + ln( n 1) 2 ln n = ln( n 1) ln n ln n + ln( n + 1) = ln n 1 n [ln n ln( n + 1)] = ln n 1 n ln n n + 1 . Let s k = k S n =2 ln · 1 1 n 2 ¸ = k S n =2 · ln n 1 n ln n n + 1 ¸ for k 2 . Then s k = · ln 1 2 ln 2 3 ¸ + · ln 2 3 ln 3 4 ¸ + · · · + · ln k 1 k ln k k + 1 ¸ = ln 1 2 ln k k + 1 , so S n =2 ln · 1 1 n 2 ¸ = lim k →∞ s k = lim k →∞ · ln 1 2 ln k k + 1 ¸ = ln 1 2 ln 1 = ln 1 ln 2 ln 1 = ln 2 . 13. (a) The x -intercepts of the curve occur where sin x = 0 x = , n an integer. So using the formula for disks (and either a CAS or sin 2 x = 1 2 (1 cos 2 x ) and Formula 99 to evaluate the integral), the volume of the n th bead is V n = π U ( n 1) π ( e x/ 10 sin x ) 2 dx = π U ( n 1) π e x/ 5 sin 2 x dx = 250 π 101 ( e ( n 1) π/ 5 e nπ/ 5 )
102 ¤ PROBLEMS PLUS (b) The total volume is π U 0 e x/ 5 sin 2 x dx = S n =1 V n = 250 π 101 S n =1 [ e ( n 1) π/ 5 e nπ/ 5 ] = 250 π 101 [telescoping sum]. Another method: If the volume in part (a) has been written as V n = 250 π 101 e nπ/ 5 ( e π/ 5 1) , then we recognize S n =1 V n as a geometric series with a = 250 π 101 (1 e π/ 5 ) and r = e π/ 5 .

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