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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 Taylor’s theorem (which is the same as the mean value theorem in this case), sin(1 /n ) = sin(0) + sin ( ξ ) 1 n for some 0 < ξ < 1 /n . So  sin(1 /n )  =  cos( ξ )  1 n ≤ 1 n . So  sin(1 /n ) n α  ≤ 1 n α +1 . Since α + 1 > 1, then ∑ 1 n α +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
 Calculus, Intermediate Value Theorem, Mean Value Theorem, Continuous function, Metric space, Compact space

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