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Test 2a Solutions

# Test 2a Solutions - n →∞ a n = c exists we have that c...

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Solution to the Second Test C1 I: a) True, b) True, c) False (Example a n = 1 /n 2 , b n = n .), d) 25, e) ln(exp(ln(1 / 4)))+ ln(exp(ln(4))) = ln(1 / 4) + ln(4) = ln(1) = 0. II: a) 4 / 2, b) 1 / 2, c) 1, d) (1 - 2 /n - 15 /n 2 ) = (1 - 5 /n )(1 + 3 /n ) and hence lim n →∞ (1 - 2 /n - 15 /n 2 ) n = lim n →∞ (1 - 5 /n ) n lim n →∞ (1 + 3 /n ) n = e - 5 e 3 = e - 2 . e) As a first guess for n large (1 + 1 n ) n 2 e n and hence one expects that the limit does not exist. This is a good argument. Another one is to use Bernoulli’s inequality (1 + 1 n ) n 2 1 + n 2 1 n = 1 + n which also shows that the limit does not exist. III: a) Clarly 0 a n 1 since n/ ( n + 1) 1 and a n = 1. b) The sequence is decreasing. c) The sequence is decreasing and bounded below by zero and hence it converges. d) a n = n ( n - 1)( n - 2) · · · 1 ( n + 1) n ( n - 1) · · · 2 = 1 n + 1 and hence lim n → ∞ a n = 0 . e) Stopping rule: Find N ( ε ) so that a n < ε whenever n > N ( ε ) Thus, 1 N ( ε ) + 1 < ε and hence N ( ε ) is the smallest integer greater than 1 . III:

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Unformatted text preview: n →∞ a n = c exists we have that c = c + 90 10 or c = 10. Thus we expect that the sequence increases towards 10. a) The sequence is bounded above by 10. To see this we use induction. It is true ±or a 1 since a 1 = 5 < 10. Assume it is true ±or some n , i.e., a n ≤ 10. Then a n +1 ≤ 10 + 90 10 = 10 1 which proves what we have claimed. b) Next we check that a n is increasing. Calculate a n +1-a n = a n + 90 10-a n = 9 10-a n 10 ≥ since a n ≤ 10. c) c = 10. d) Next, note that ≤ 10-a n +1 = 10-a n + 90 10 = 10-a n 10 = 1 10 (10-a n ) . Thus ≤ 10-a n +1 = ± 1 10 ² n (10-a 1 ) = 5 ± 1 10 ² n . Thus if n = 7 we get that 510-7 < 10-6 and hence a n < 10-6 for all n ≥ 8. 2...
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Test 2a Solutions - n →∞ a n = c exists we have that c...

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