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solutionshw7-554-2010

solutionshw7-554-2010 - ≤ b n frequently If a> b then...

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Solutions homework 7. Page 39 Problem 3. : False. Take ( x n ) = (1 , 0 , 1 , 0 , · · · ) and ( y n ) = (0 , 1 , 0 , 1 , · · · ). Then neither ( x n ) or ( y n ) is a null sequence, but their product is identical zero. Page 39 Problem 4. True. Note x n = ( n + 1 - n ) · n + 1 + n n + 1 + n = 1 n + 1 + n 0 . Page 39 Problem 6. From Calculus II we know that cos x = 1 - x 2 2 + x 4 4! - · · · . Hence 1 - cos( 1 n ) = 1 2 n 2 - 1 24 n 4 + · · · . It follows that x n = 1 2 n - 1 24 n 3 + · · · ≈ 1 2 n , which is a null sequence. Page 39 Problem 8. We use the same trick as in problem 4, noting that n = n 2 . Hence a n = ( n 2 + 1 - n 2 ) · n 2 + 1 + n 2 n 2 + 1 + n 2 = 1 n 2 + 1 + n 2 0 . Page 41 Problem 1. We first observe that if x n x and x > 0, then x n > 0 ultimately. The proof of this is very similar to the proof of Theorem 3.4.8 (6): Let r = x 2 . Then there exists n such that | x - x n | < r for all n N . Now 2 r = x = x - x n + x n ≤ | x - x n | + x n < r + x n for all n N implies that x n r > 0 for all n N . Now assume a n
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Unformatted text preview: ≤ b n frequently. If a > b , then a-b > 0 and a n-b n → a-b implies by the first part that ultimately ( a n-b n ) > 0, which contradicts a n ≤ b n frequently. Page 41 Problem 7. Observe that a n +1-a n = 1 2 n +1 + 1 2( n +1)-1 n +1 = 1 2 n +1-1 2( n +1) > 0, so ( a n ) is increasing. To bound a n from above, note that a n is a sum of n terms each of which is less or equal to 1 n +1 , so a n ≤ n n +1 < 1. Hence ( a n ) is increasing and bounded above, and thus convergent (its limit is the supremum). Page 47 Problem 3. True, since | sin n | ≤ 1 for all n , we have a bounded sequence, which by the Bolzano-Weierstrass theorem has a convergent subsequence. 1...
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