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Unformatted text preview: sums. Let P n = { , 1 /n,. .., ( n1) /n, 1 } . To form a Riemann sum with respect to this partition, we choose a point in each interval [( i1) /n,i/n ], namely i/n . Then we observe that the lefthand side is equal to lim n S ( P n ,f ) . 1 Since lim n kP n k = 0, the above must tend to R 1 f ( x ) dx , as desired. Exercise 7.2.8 Suppose that a k converges. Prove that lim n 1 n n X k =1 ka k = 0 . Proof : Using the Abel partial summation formula (Theorem 7.2.1) for p = 1, q = n and b k = k , we get n X k =1 ka k = nA nn1 X k =1 A k . Therefore, lim n 1 n n X k =1 ka k = lim n ( A n1 n n1 X k =1 A k ) = AA = 0 , by exercise 2.2.14. 2...
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 Winter '08
 hitrik
 Real Numbers

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