m147A2sol - MATH 147 Assignment #2 1) For each of the...

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Unformatted text preview: MATH 147 Assignment #2 1) For each of the following sequences determine if it converges or diverges. If it converges find the limit: a) { n n +1 } b) { sin( n ) n } c) { n + 1- n } d) e n n ! e) { n cos( n ) 2 n +1 } Solutions: [25 marks (5 marks each)] a) Since { n n +1 } = 1- 1 n +1 and lim n 1 n +1 = 0, we have that lim n n n + 1 = 1- lim n 1 n + 1 = 1 b) We know that | sin( n ) | 1 for each n . This means that | sin( n ) n | 1 n . This gives us- 1 n sin( n ) n 1 n Since lim n - 1 n = 0 = lim n 1 n , we can apply the Squeeze Theorem to show that lim n sin( n ) n = 0. c) Observe that lim n n + 1- n = lim n ( n + 1- n )( n + 1 + n n + 1 + n ) = lim n 1 n + 1 + n = 0 . d) First observe that if | r | < 1, then lim n r n = 0 (why?). Now for n 3, e n n ! = e 1 e 2 e 3 e 4 e n 1 e 1 e 2 e 3 e 3 e 3 = e 2 2 ( e 3 ) n- 2 We have shown that for n 3 e n n ! e 2 2 ( e 3 ) n- 2 Since e 3 < 1, we have lim n 0 = 0 = lim n e 2 2 ( e 3 ) n- 2 The Squeeze Theorem shows that lim n e n n ! = 0. e) Observe that if a n = n cos( n ) 2 n +1 , then lim k a 2 k = lim k 2 k cos(2 k ) 2(2 k ) + 1 = lim k 2 k 4 k + 1 = 1 2 while lim k a 2 k- 1 = lim k (2 k- 1) cos((2 k- 1) ) 2(2 k- 1) + 1 = lim k - 2 k + 1 4 k- 1 =- 1 2 Since a convergent sequence cannot have subsequences converging to different values this sequence diverges. 2) We will later be able to show that 1 n + 1 < ln( n + 1)- ln( n ) < 1 n 2 a) Let a n = 1 + 1 2 + 1 3 + + 1 n- ln( n ). Prove that { a n } converges. Note: = lim n a n is called Eulers constant. It is still not known whether or not is rational. Solution: [5 marks] Observe that a n +1- a n = 1 n +1- (ln( n +1)- ln ( n )) < 0 so the sequence is decreasing. Now part b) below shows that a n = 1 + 1 2 + 1 3 + + 1 n- ln( n ) 1 + 1 2 + 1 3 + + 1 n- ln( n + 1) > We can now use the MCT to show that the sequence converges. b) Show that if b n = 1 + 1 2 + 1 3 + + 1 n , then ln( n + 1) < b n ln( n ) + 1. Solution: [5 marks] The given inequality shows that ln n + 1 = n X i =1 (ln( i + 1)- ln( i )) < n i =1 1 i = b n It also shows that for n 2, b n- 1 = n X i =2 1 n < n X i =2 (ln( n )- ln( n- 1)) = ln( n ) Hence for n 2, we have that b n < ln( n ) + 1. When n = 1, we get that b 1 = 1 = ln (1) + 1. It follows that ln( n + 1) < b n ln( n ) + 1 for each n . 3 c) Estimate how many terms it would take before b n > 10 6 . How long do you think it would it take for a modern computer to perform this many additions?...
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This note was uploaded on 10/21/2010 for the course MATH 147 taught by Professor Wolzcuk during the Fall '09 term at Waterloo.

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m147A2sol - MATH 147 Assignment #2 1) For each of the...

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