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 δ > 0 as in the hint. Then take 1 n < δ and take σ = { < 1 n < ··· < k n < ··· < n n = 1 } . Then s ( σ ) ≤ 1 n ∑ n k =1 f ( k n ) ≤ S ( σ ) and S ( σ )s ( σ ) < ± . 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 ² = 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 ² = 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 = π 4 . 1...
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 Spring '07
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 Geometric Series, lim, 1 K, 1 k, bn converges

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