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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.
3
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
1,2
REFERENCES
Ali Shadhar Olaikhan, An Introduction to the Harmonic Series and Logarithmic Integrals, 2021, p. 284, eq. (4.209).
LINKS
Philippe Flajolet and Bruno Salvy, Euler Sums and Contour Integral Representations, Experimental Mathematics 7:1 (1998), pp. 15-35. See page 23.
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)
Equals Sum_{k>=1} H(2*k) * AH(2*k) / k^2, where H(k) = A001008(k)/A002805(k) is the k-th harmonic number, and AH(K) = A058313(k)/A058312(k) is the k-th alternating harmonic (or skew-harmonic) number (Olaikhan, 2021). - Amiram Eldar, Feb 03 2026
EXAMPLE
1.894065658994491835153006468947043829855814165857772088445208377027211...
MATHEMATICA
RealDigits[7/4*Zeta[4], 10, 100] // First
PROG
(PARI) 7*zeta(4)/4 \\ Michel Marcus, Jul 04 2014
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
STATUS
approved