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# HW12 - = 0 A-B = 6 ⇒ 2 A = 6 ⇒ A = 3 and B =-3 Hence k...

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Math 191 HW 12 Solutions 11.1: 16 a n = ( - 1) n +1 n 2 , n = 1 , , 2 , ... 11.1: 32 lim n →∞ ( - 1) n (1 - 1 n ) does not exist diverges. 11.1: 50 lim n →∞ 1 - 1 n n = lim n →∞ 1 + ( - 1) n n = e - 1 converges. (Theorem 5, #5) 11.1: 82 lim n →∞ 1 n 2 - 1 - n 2 + n = lim n →∞ ( 1 n 2 - 1 - n 2 + n )( n 2 - 1 + n 2 + n n 2 - 1 - n 2 + n ) = lim n →∞ n 2 - 1 + n 2 + n 1 - n = lim n →∞ q 1 - 1 n 2 + q 1 + 1 n - 1 n - 1 = - 2 converges. 11.2: 6 5 n ( n +1) = 5 n - 5 n +1 S n = (5 - 5 2 )+( 5 2 - 5 3 )+( 5 3 - 5 4 )+ ... +( 5 n - 1 - 5 n )+( 5 n - 5 n +1 ) = 5 - 5 n +1 lim n →∞ S n = 5 11.2: 12 (5 - 1) + ( 5 2 - 1 3 ) + ( 5 4 - 1 9 ) + ( 5 8 - 1 27 ) + ..., is the difference of two geometric series; the sum is 5 1 - ( 1 2 ) - 1 1 - ( 1 3 ) = 10 - 3 2 = 17 2 11.2: 14 2 + 4 5 + 8 25 + 16 125 + ... = 2(1 + 2 5 + 4 25 + 8 125 + ... ); the sum of this geometric series is 2 1 1 - ( 2 5 ) = 10 3 11.2: 16 6 (2 n - 1)(2 n +1) = A 2 n - 1 + B 2 n +1 = A (2 n +1)+ B (2 n - 1) (2 n - 1)(2 n +1) A (2 n + 1) + B (2 n - 1) = 6 (2 A + 2 B ) n
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Unformatted text preview: = 0 A-B = 6 ⇒ 2 A = 6 ⇒ A = 3 and B =-3. Hence, k X n =1 6 (2 n-1)(2 n + 1) = 3 k X n =1 1 2 n-1 + 1 2 n + 1 = 3 ± 1 1-1 3 + 1 3-1 5 + 1 5-1 7 + ...-1 2( k-1) + 1 + 1 2 k-1-1 2 k + 1 ² = 3(1-1 2 k + 1 ) ⇒ the sum is lim k →∞ 3(1-1 2 k + 1 ) = 3. 11.2: 30 lim n →∞ ln 1 n =-∞ 6 = 0 ⇒ diverges. 11.2: 34 lim n →∞ ± 1-1 n ² n = lim n →∞ ± 1 +-1 n ² n = e-1 6 = 0 ⇒ diverges. 11.2: 48 a = 1 , r = 3-x 2 ; converges to 1 1-( 3-x 2 ) = 2 x-1 for | 3-x 2 < 1 or 1 < x < 5. 1...
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