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Re: Puzzled by Sum


Yes, but:

A[i_, j_] := i/j + j/i
F[X_] := Module[{i, j}, Sum[A[X - j, j], {j, 1, X - 1}]]

F[X] /. X -> 2
F[i] /. i -> 2

2

-1 - 2 DifferenceRoot[
    Function[{\[FormalY], \[FormalN]}, {(-2 + \[FormalN]) \[FormalY][\
\[FormalN]] + (3 - 2 \[FormalN]) \[FormalY][
          1 + \[FormalN]] + (-1 + \[FormalN]) \[FormalY][
          2 + \[FormalN]] == 0, \[FormalY][0] == 
       0, \[FormalY][1] == -(1/2)}]][2]



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