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Unformatted text preview: lim n →∞ √ n + 11 = ∞ This series, however is conditionally convergent since b n +1 < b n and that lim n →∞  b n  = 0. Problem 9 (4 marks) ∞ X n =1 sin ( n ) √ n n 2 + 1 This series is absolutely convergent (and therefore conditionally convergent) since ∞ X n =1  sin ( n ) √ n n 2 + 1 ≤ ∞ X n =1 √ n n 2 = ∞ X n =1 1 n 3 2 This series is convergent from the pseries test. Problem 11 (4 marks) ∞ X n =1 sin ( 1 n ) ( ln (1 + n )) 2 We can see that this is convergent by using the comparison test using the following convergent series ∞ X n =1  sin ( 1 n ) ( ln (1 + n )) 2 ≤ ∞ X n =1 1 n ( ln ( n )) 2 This series can be shown to be convergent by the integral test Z ∞ 2 1 x ( ln ( x )) 2 dx = 1 ln ( x ) ± ± ± ± 2 ∞ = 1 ln (2 < ∞ Therefore the series is conditionally convergent 2...
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 Winter '08
 KarlPeterRussell
 Math, Calculus, Sin, Mathematical Series, lim, n=1

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