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Re: Walker's Equation

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  • djbroadhurst
    PS: The proof is elementary sum_{n 1,m 1} 1/n^m = sum_{n 1} 1/(n*(n-1)) = sum_{n 1} (1/(n-1) - 1/n) = 1/(2-1) = 1 Numerical check: ? p60 realprecision = 67
    Message 1 of 3 , Feb 2, 2002
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      PS: The proof is elementary

      sum_{n>1,m>1} 1/n^m = sum_{n>1} 1/(n*(n-1))
      = sum_{n>1} (1/(n-1) - 1/n) = 1/(2-1) = 1

      Numerical check:

      ? \p60
      realprecision = 67 significant digits (60 digits displayed)
      ? print(suminf(n=2,zeta(n)-1))
      1.00000000000000000000000000000000000000000000000000000000000
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