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Unformatted text preview: 18.014 Problem Set 11 Solutions Total: 32 points Problem 1: Prove that a sequence converges if, and only if its liminf equals its particular, the set { a n } for n > N is bounded below by L − and bounded above limsup. Solution (4 points) Suppose { a n } is a sequence that converges to a limit L . Then given > 0, there exists an integer N such that n > N implies  a n − L  < 2 . In 2 by L + 2 . Thus, L + inf a n and L + sup a n . 2 ≥ n ≥ N +1 ≥ L − 2 2 ≥ n ≥ N +1 ≥ L − 2 In particular, given > 0, there exists N such that m > N implies < , a n − L < . inf a n − L sup n ≥ m n ≥ m We conclude lim inf a n = L = lim sup a n and the liminf equals the limsup. Conversely, suppose { a n } is a sequence such that lim inf a n = L = lim sup a n . We will show lim n →∞ a n = L . Given > 0, there exist N 1 and N 2 such that m > N 1 implies inf a n − L < n ≥ m and m > N 2 implies sup n ≥ m a n − L < . Let N = max { N 1 ,N 2 } . If m > N 1 ,N 2 , then L − < inf a n ≤ a m ≤ sup a n < L + . n ≥ m n ≥ m In particular,  a m − L  < if m > N . Thus, lim n →∞ a n = L . 1   Problem 2: Use this fact to prove every Cauchy sequence of real numbers converges. Solution (4 points) First, recall the following lemma: Lemma: Every decreasing sequence that is bounded below converges, and every increasing sequence that is bounded above converges. Now, let { a n } be a Cauchy sequence of real numbers. Then there exists M such that n,m > M implies  a n − a m  < 1. Putting m = M + 1, we observe a M +1 − 1 ≤ a n ≤ a M +1 + 1 if n > M . Put C = max { a M +1 ,a 1 ,...,a M } and put B = min { a M +1 − 1 ,a 1 ,...,a M } . Then B ≤ a n ≤ C for all n . In particular, if b n = sup a m and c n = inf m ≥ n a m , then { b n } is a de m ≥ n creasing sequence bounded below by B and { c n } is an increasing sequence bounded above by C . Thus, by the lemma { b n } converges to L 1 and { c n } converges to L 2 ....
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This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.
 Fall '10
 ChristineBreiner
 Calculus

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