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EXAMPLE
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The algorithm at A226049, with r = 2 and f(n) = n^(-1/2), gives
1/1 + 1/sqrt(2) + 1/sqrt(3) - 1/sqrt(12) + 1/sqrt(56204) - ... ,
converging to 2. The 11th partial sum differs from 2 by less than 10^(-19000).
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MATHEMATICA
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$MaxExtraPrecision = Infinity; z = 10; f[n_] := n^(-1/2); g[n_] := 1/n^2; r = 2; s = 0; a[1] = NestWhile[# + 1 &, 1, ! (s += f[#]) >= r &]; p = Sum[f[n], {n, 1, a[1]}]; a[2] = Floor[g[p - r]]; a[n_] := Floor[g[((-1)^n) (p - r - Sum[((-1)^k) f[a[k]], {k, 2, n - 1}])]];
Table[a[k], {k, 1, z}]
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