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Unformatted text preview: b n converges, then ( t n ) is bounded, which implies that ( s n ) is bounded and thus a n converges whenever b n converges. Extra Problem 1. Let > 0 and take > 0 as in the hint. Then take 1 n < and take = { < 1 n < < k n < < n n = 1 } . Then s ( ) 1 n n k =1 f ( k n ) S ( ) and S ( )s ( ) < . Hence  1 n n k =1 f ( k n )R 1 f  < . Extra Problem 2. a. Take f ( x ) = x 2 . Then 1 n n X k =1 f k n = 1 n 3 n X k =1 k 2 , so lim n 1 n 3 n X k =1 k 2 = Z 1 x 2 dx = 1 3 . b. Take f ( x ) = 1 1+ x 2 . Then 1 n n X k =1 f k n = n X k =1 n n 2 + k 2 , so lim n n X k =1 n n 2 + k 2 = Z 1 1 1 + x 2 dx = arctan x  1 = 4 . 1...
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This note was uploaded on 02/05/2012 for the course MATH 555 taught by Professor Staff during the Spring '07 term at South Carolina.
 Spring '07
 Staff
 Geometric Series

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