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Unformatted text preview: 18.100 Midterm #2 Solutions Problem 1 (a) The k th partial sum k n =0 ( a n a n +1 ) is equal to a a k +1 . So lim k X n =0 ( a n a n +1 ) = lim k a a k +1 = a . (b) By Taylors theorem (which is the same as the mean value theorem in this case), sin(1 /n ) = sin(0) + sin ( ) 1 n for some 0 < < 1 /n . So  sin(1 /n )  =  cos( )  1 n 1 n . So  sin(1 /n ) n  1 n +1 . Since + 1 > 1, then 1 n +1 converges and by the comparison test, so does n =1 sin(1 /n ) n . Problem 2 (a) We prove this by induction. We have x = 2 > 2. Assume x n > 2. Then x n +1 = 1 2 x n + 2 x n > 2 x 2 n + 2 > 2 2 x n x 2 n 2 2 x n + 2 > ( x n 2) 2 > 0. The latter statement is true since x n > 2 by the inductive hypothesis. (b) We need to show that x n +1 = 1 2 x n + 2 x n x n x 2 n + 2 2 x 2 n x 2 n 2 which is true from (a). Since x n is monotonically decreasing and bounded from below, then lim...
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 Fall '10
 Prof.KatrinWehrheim
 Mean Value Theorem

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