q3s - to nd the limit lim n s n . By part (a) the sequence...

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Math 310 - Quiz 2 Solutions Monday, 19 Oct 2009 No calculators, notes or text allowed. Let ( s n ) n be a sequence defined recursively by s 1 = π and s n +1 = 2 - 1 s n for n N . a. Use induction to prove that s n 1 for all n N . For the base case observe that s 1 = π > 1. Now, suppose that s n 1 for some n N . Then, we have 1 s n 1 and so s n +1 = 2 - 1 s n 2 - 1 = 1 . Hence, by induction, s n 1 for all n N . ± b. Show that the sequence ( s n ) n is nonincreasing (you may use the fact that x + 1 x 2 for all real x > 0). Note that the given inequality implies that 2 - 1 x x for all real x > 0. Let n N . Then since s n 1 > 0 we have s n +1 = 2 - 1 s n s n . Therefore, the sequence ( s n ) n is nonincreasing as desired. ± c. Prove that ( s n ) n converges (show that the theorem on bounded monotone sequences applies); use limit theorems
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Unformatted text preview: to nd the limit lim n s n . By part (a) the sequence is bounded below and by part (b) the sequence is nonincreasing. Hence, by the Monotone Convergence Theorem, the sequence ( s n ) n converges. Let s denote the limit. We have by various algebraic limit theorems that s = lim n s n +1 = lim n 2-1 s n = lim n 2-lim n 1 s n = 2-1 s . Hence, s = 2-1 s and so 0 = s 2-2 s + 1 = ( s-1) 2 . It follows that lim n s n = 1....
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