117-022409-1

117-022409-1 - MATH 117 LECTURE NOTES FEBRUARY 24, 2009 DR....

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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. Lets 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 }} . Lets call the set of subsequential limits S. So, weve shown that lim sup s n = X sup S. To show that theyre 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 FEBRUARY 24, 2009 DR....

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