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

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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( ) 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
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e 1 e 2 e 3 e 3 · · · e 3 = e 2 2 ( e 3 ) n - 2 We have shown that for n 3 0 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( ) 2 n +1 , then lim k →∞ a 2 k = lim k →∞ 2 k cos(2 ) 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) π ) 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
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a) Let a n = 1 + 1 2 + 1 3 + · · · + 1 n - ln( n ). Prove that { a n } converges. Note: γ = lim n →∞ a n is called Euler’s constant. It is still not known whether or not γ 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) > 0 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
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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? Solution: [5 marks] For b n > 10 6 it would suffice to have ln( n + 1) > 10 6 or n > e 10 6 - 1 terms. This is an immense number of additions to perform. No modern computer would survive long enough to perform anywhere near this many operations. 3) a) Suppose that a n 0 and that lim n →∞ a n = L . Show that lim n →∞ a n = L . (Hint: Do the cases L = 0 and L > 0 separately. When L > 0 show that a n - L = a n - L a n + L .
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