222A1-solutions

222A1-solutions - lim n n + 1-1 = This series, however is...

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Math 222: Assignment 1 Problem 1 (2 marks) Does the sequence b n = sin ( π 2 + 1 n ) converge or diverge. If it converges what is the limit? lim n →∞ sin ( π 2 + 1 n ) = sin ( π 2 + lim n →∞ 1 n ) = sin ( π 2 ) = 1 Problem 3 (2 marks) Does the sequence b n = n 2 (1 - cos ( 1 n )) converge or diverge. If it converges what is the limit? lim n →∞ n 2 (1 - cos ( 1 n )) = lim n →∞ (1 - cos ( 1 n )) 1 n 2 Applying L’Hopitals Rule twice we get lim n →∞ (1 - cos ( 1 n )) 1 n 2 = lim n →∞ sin ( 1 n ) 2 = lim n →∞ 1 2 cos ( 1 n ) So we get the limit of b n to be 1 2 . For the following series, determine whether they converge (absolutely or conditionally) or diverge. Justify your answer. Problem 5 (4 marks) X n =1 ( - 1) n +1 1 + n n 2 First we note that X n =1 | ( - 1) n +1 1 + n n 2 |≤ X n =1 1 n + X n =1 1 n 2 And so is divergent because the harmonic series diverges. So for conditional convergence we notice that b n +1 < b n and that lim n →∞ | b n | = 0. So we have conditional convergence. 1
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Problem 7 (4 marks) X n =1 ( - 1) n +1 1 n + 1 + n We do not have absolute convergence since X n =1 1 n + 1 + n = X n =1 n + 1 - n This is a telescoping sum of which the limit is
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Unformatted text preview: lim n n + 1-1 = This series, however is conditionally convergent since b n +1 &lt; 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 p-series 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 &lt; Therefore the series is conditionally convergent 2...
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222A1-solutions - lim n n + 1-1 = This series, however is...

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