Geometric Series Sequences and L Hopital s Rule soln

Thomas' Calculus: Early Transcendentals

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Mat104 Solutions to Problems from Old Exams Geometric Series, Sequences and L’Hˆopital’s Rule (1) Since e nx = ( e x ) n this is a geometric series with r = e x . It converges absolutely, provided | e x | < 1, that is for x ( -∞ , 0). In that case, it will converge to 1 1 - e x . (2) This is a geometric series with a = ( - 2 / 3) 4 and r = - 2 / 3. Therefore it converges to ( - 2 / 3) 4 1 + 2 / 3 = 16 135 . (3) Here we combine several geometric series: X n =0 2 n 5 n = 1 1 - 2 / 5 = 5 / 3 X n =0 3 n +1 5 n = 3 1 - 3 / 5 = 15 / 2 X n =0 4 n +2 5 n = 16 1 - 4 / 5 = 80 the series we are given will converge to 5 / 3 + 15 / 2 + 80 = . . . . (4) Answer: 2 + 1 / 2 - 3 / 8 = 17 / 8 (similar to problems 1-3 above) (5) Answer: 8 / 3 + 2 = 14 / 3. (similar to problems 1-3 above) (6) As n → ∞ , both the numerator and the denominator go to infinity. Thus we can use L’Hˆopital’s Rule: lim n →∞ ln( n 2 + n ) ln( n 2 - n ) = lim
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Unformatted text preview: n →∞ 2 n + 1 n 2 + n 2 n-1 n 2-n = lim n →∞ 2 n + 1 2 n-1 · n 2-n n 2 + n = 1 since the leading term on top and bottom is now 2 n 3 . (7) This limit will be 0. If we apply L’Hˆopital’s Rule repeatedly we end up with 18 24 n which goes to 0 as n goes to infinity. (8) This limit exists and equals-1. (Recall arctan n goes to π/ 2 as n goes to infinity.) (9) Since (1 + 1 /n ) n goes to e as n goes to infinity, this will go to e 2 . (10) (a) The numerator goes to e and the denominator goes to ∞ . So the quotient will go to 0 as n goes to ∞ . (b) An ∞ ∞ form so use L’Hˆ opital’s Rule to show that the limit is 1 / 2. 1...
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