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# notes 1 - The sum in problem 6.4a The sum we need to...

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The sum in problem 6.4a The sum we need to complete is, S = 1 n 2 m 2 m n . This can be simplified using partial fractions, S = 1 2 n 1 n m + 1 n + m m n = 1 2 n 1 n m m n + 1 n + m m n . The sums can each be broken into the part up to n-1 and the part from n+1 upward, S = 1 2 n 1 n m m = 0 n 1 + 1 n m m = n + 1 k + 1 n + m m = 0 n 1 + 1 n + m m = n + 1 k , where k will be allow to approach infinity shortly. Reordering the sums, S = 1 2 n 1 n m m = 0 n 1 + 1 n + m m = 0 n 1 + 1 n + m m = n + 1 k + 1 n m m = n + 1 k . Replacing m with –m in the first sum then flipping the limits, S = 1 2 n 1 n + m m = 1 n 0 + 1 n + m m = 0 n 1 + 1 n + m m = n + 1 k + 1 n m m = n + 1 k Now, the first three sums can be turned into a single sum if we pull out one of the redundant m=0 terms ( 1 n ) from the first two sums and subtract off the m=n term ( 1 2 n ) that needs to be added write a single sum, S = 1 2 n 1 n 1 2 n + 1 n + m m = 1 n k + 1 n m m = n + 1 k . Now, replace m with j-n in the first sum, S = 1 2 n 1 n 1 2 n
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