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solutionshw7-555-2011

solutionshw7-555-2011 - ∑ b n converges then t n is...

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Solutions homework 7. Page 182 Problem 1. The answer is 2 times the sum of the geometric series with c = 1 3 , so the answer is 2 · 1 1 - 1 3 = 3. Page 182 Problem 4. For n 1 we have s 2 n +2 = s 2 n + a 2 n +1 + a 2 n +2 s 2 n (since a 2 n +1 + a 2 n +2 0), so ( s 2 n ) is an increasing sequence. Similarly s 2 n +1 = s 2 n - 1 + a 2 n + a 2 n +1 s 2 n - 1 , so ( s 2 n - 1 ) is a decreasing sequence. Moreover s 2 n +1 = s 2 n + a 2 n +1 s 2 n , From this it follows that ( s 2 n ) is bounded above by s 1 and that ( s 2 n +1 ) is bounded below by s 2 . Hence both limits lim s 2 n and lim s 2 n +1 exist. Now s 2 n +1 = s 2 n + a 2 n +1 and a n +1 0 show that these two limits are equal. Hence lim s n exists and the series converges. Page 184 Problem 1. If ( s n ) is bounded, then by Bolzano–Weierstrass this sequence gas a convergent subsequence ( s n k ). Using these n 0 k s for the grouping, we get as in the proof of Theorem 10.2.3 that ( t k ) = ( s n k ) converges. Page 187 Problem 2. Let s n = a 1 + ··· + a n and t n = b 1 + ··· + b n . Then a n converges if and only if ( s n ) is bounded, and b n converges if and only if ( t n ) is bounded, since both series are positive term series. Hence If
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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 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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