Lecture39-rev

# Finally if r 1 then the sequence ar k is a a a a

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Finally if r = - 1 then the sequence { ar k } is - a, a, - a, a, · · · which doesn’t converge (since a = 0). Theorem 3 (Sandwich Theorem (or Squeeze Law)). Let a k , b k , and c k be three sequences satisfying a k b k c k for k = 1 , 2 , . . . . Then lim k →∞ a k = L and lim k →∞ c k = L = lim k →∞ b k = L 1.0.3 Example Calculate lim k →∞ k - 2 sin k . SOLUTION: Since - 1 sin k 1 for every k we have - k - 2 k - 2 sin k k - 2 Now by the sandwich theorem, since lim k →∞ - k - 2 = lim k →∞ k - 2 = 0 we get lim k →∞ k - 2 sin k = 0 2 Bounded Monotone Convergence Theorem So far we have had no result guaranteeing the existence of a given limit with- out deriving it from the existence of another limit, or using a bare hands ap- proach starting with the definition of the limit. This lack is remedied by the bounded monotone convergence theorem . We begin with the meanings of the terms monotone and bounded 2

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2.1 Definition A sequence a 1 , a 2 , . . . with general term a k is called increasing if a k < a k +1 for all k decreasing if a k > a k +1 for all k monotone if it is either increasing or decreasing. 2.2 Example Let a k = k +3 k +2 . Is this sequence monotone? First Solution. a k +1 - a k = k +4 k +3 - k +3 k +2 = ( k +4)( k +2) - ( k +3) 2 ( k +3)( k +2) = - 1 ( k +3)( k +2) < 0. Thus a k +1 < a k for all k and so the sequence is decreasing and therefore monotone.
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