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117-022409-1

117-022409-1 - MATH 117 LECTURE NOTES DR JULIE ROWLETT 1...

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Unformatted text preview: MATH 117 LECTURE NOTES FEBRUARY 24, 2009 DR. JULIE ROWLETT 1. Limit Superior and Limit Inferior Theorems Theorem 1.1. For a bounded sequence of real numbers { s n } the limit superior is equal to the supremum of all subsequential limits of { s n } . Proof: Let the lim sup be X. Let’s define X n := sup k ≥ n s k . Then, X = lim n →∞ X n . By definition of limit, for any > , there exists N ∈ N such that for n ≥ N, | X n- X | < 2 . Note that X n ≥ s k for all k ≥ n, which shows that X n ≥ L for any subsequential limit L. (why?) Hence, the same is true for the limit, by Theorem 17.4, so that X ≥ L- 2 . Since > 0 was arbitrary, this shows that X ≥ L for any subsequential limit L. Hence, X is an upper bound for the set of subsequential limits. So, by definition of supremum, X ≥ sup { subsequential limits of { s n }} . Let’s call the set of subsequential limits S. So, we’ve shown that lim sup s n = X ≥ sup S. To show that they’re equal, we must now show that X ≤ sup S. By definition of supremum, for > , X n- 2 is not an upper bound for { s k } k ≥ n , so there exists s n k with n k ≥ n such that X n- 2 < s n k ≤ X n . By the triangle inequality, then | s n k- X | ≤ | s n k- X n | + | X n- X | < . Now, let m = 2- m . Then, there exists s n 1 such that | s n 1- X | ≤ 1 = 2- 1 . By our preceding arguments, for each m, there exists s n m such that n m- 1 < n m and | s n m- X | ≤ e m = 2- m ....
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117-022409-1 - MATH 117 LECTURE NOTES DR JULIE ROWLETT 1...

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