A244664 Decimal expansion of Sum_{n >= 1} H(n,2)/n^2 where H(n,2) = A007406(n)/A007407(n) is the n-th harmonic number of order 2.
1, 8, 9, 4, 0, 6, 5, 6, 5, 8, 9, 9, 4, 4, 9, 1, 8, 3, 5, 1, 5, 3, 0, 0, 6, 4, 6, 8, 9, 4, 7, 0, 4, 3, 8, 2, 9, 8, 5, 5, 8, 1, 4, 1, 6, 5, 8, 5, 7, 7, 7, 2, 0, 8, 8, 4, 4, 5, 2, 0, 8, 3, 7, 7, 0, 2, 7, 2, 1, 1, 0, 7, 8, 3, 2, 7, 1, 9, 5, 4, 8, 1, 4, 7, 4, 5, 9, 1, 8, 6, 2, 8, 9, 7, 9, 7, 4, 8, 5, 5
Offset: 1
Examples
1.894065658994491835153006468947043829855814165857772088445208377027211...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Philippe Flajolet, Bruno Salvy, Euler Sums and Contour Integral Representations, Experimental Mathematics 7:1 (1998) page 23.
Programs
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Mathematica
RealDigits[7/4*Zeta[4], 10, 100] // First
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PARI
7*zeta(4)/4 \\ Michel Marcus, Jul 04 2014
Formula
Equals 7*Pi^4/360 = (7/4)*A013662.
From Peter Bala, Jul 27 2025: (Start)
Series acceleration formula:
Let s(n) = Sum_{k = 1..n} H(k,2)/k^2 and S(n) = Sum_{k = 1..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*s(n+k). It appears that S(n) converges much more rapidly to 7*Pi^4/360 than s(n).
For example, s(50) = 1.8(61...) is only correct to 2 decimal digits, while S(50) = 1.89406565899449183515 30064689470(06...) is correct to 32 decimal digits. (End)