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A262031
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Numerator of partial sums of a reordered alternating harmonic series.
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3
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1, 4, 5, 31, 247, 389, 1307, 15637, 13327, 187111, 199123, 353201, 6364777, 127056883, 23083451, 24191987, 579694957, 535076383, 13912332463, 43224283189, 40355946289, 1210479158981, 38689398709811, 72866186391697, 75054119011297, 77117026909777, 73105817107177, 2777117009412349
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OFFSET
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0,2
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COMMENTS
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The reordered alternating harmonic series considered here is 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + 1/11 - 1/6 + ... + ... - ...
The limit n -> infinity of the partial sums s(n) = a(n)/A262031(n) is 3*log(2)/2, approximately 1.03972077083991... For the decimal expansion see A262023.
Combining three consecutive terms of this series leads to the series b(0) + b(1) + ..., with b(k) = (1/2)*(8*k+5)/((4*k+1)*(4*k+3)*(k+1)). This produces partial sums 5/6, 13/140, 7/198, 29/1560, 37/3230, ..., which are given by s(3*n+2), n = 0, 1, .... Therefore, the limit is the same as the one given above, and it is obtained from Sum_{k=0..n} b(k) = (1/4)*Psi(n+5/4) + (1/4)*Psi(n+7/4) - (1/2)*Psi(n+2) + (3/2)*log(2), with the digamma function Psi(x).
This reordered alternating harmonic series appears as an example in the famous Dirichlet article, p. 319 (Werke I). Martin Ohm showed that for the reordering with alternating m consecutive positive terms followed by n negative terms (here n = 2 and m = 1) the sum becomes log(2) + (1/2)*log(m/n). See the reference, paragraph 8. p. 12-14. See also the Pringsheim reference.
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LINKS
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FORMULA
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a(n) = numerator(s(n)) with s(n) = Sum_{k=0..n} c(k), where c(k) = 3/(4*k+3), 3/(4*k+5), -3/(2*(k+1)) if k == 0, 1, 2 (mod 3), respectively.
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EXAMPLE
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The first fractions s(n) (in lowest terms) are 1, 4/3, 5/6, 31/30, 247/210, 389/420, 1307/1260, 15637/13860, 13327/13860, 187111/180180, 199123/180180, 353201/360360, ...
The values s(10^n), for n=0..6, are (Maple 10 digits) [1.333333333, 1.105133755, 1.047114258, 1.040469694, 1.039795760, 1.039728271, 1.039721521], to be compared with 3*log(2)/2 (approximately 1.039720771).
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MATHEMATICA
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Table[Numerator@ Sum[Which[Mod[k, 3] == 0, 3/(4 k + 3), Mod[k, 3] == 1, 3/(4 k + 5), True, -3/(2 (k + 1))], {k, 0, n} ], {n, 0, 27}] (* Michael De Vlieger, Jul 26 2016 *)
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PROG
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(PARI) lista(nn) = {my(s = 0); for (k=0, nn, if (k%3==2, t = -3/(2*(k+1)), if (k%3==1, t = 3/(4*k+5), t = 3/(4*k+3))); s += t; print1(numerator(s), ", "); ); } \\ Michel Marcus, Sep 13 2015
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CROSSREFS
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KEYWORD
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nonn,frac,easy
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AUTHOR
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STATUS
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approved
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