A258982 Decimal expansion of the multiple zeta value (Euler sum) zetamult(5,3).

0, 3, 7, 7, 0, 7, 6, 7, 2, 9, 8, 4, 8, 4, 7, 5, 4, 4, 0, 1, 1, 3, 0, 4, 7, 8, 2, 2, 9, 3, 6, 5, 9, 9, 1, 4, 8, 2, 2, 6, 0, 1, 3, 1, 9, 4, 1, 5, 2, 7, 7, 5, 2, 4, 0, 1, 2, 6, 4, 5, 0, 7, 7, 8, 0, 3, 9, 1, 0, 9, 3, 8, 7, 5, 5, 5, 0, 7, 2, 1, 9, 8, 9, 1, 3, 8, 3, 6, 0, 2, 9, 8, 1, 9, 0, 7, 7, 0, 8, 6

Offset: 0

Author

Jean-François Alcover, Jun 16 2015

Keywords

• nonn
• cons

Examples

			0.03770767298484754401130478229365991482260131941527752401264507780391...
		

Links

• Richard E. Crandall, Joe P. Buhler, On the evaluation of Euler Sums, Exp. Math. 3 (4) (1994) 275-285 Table 1.
• Eric Weisstein's MathWorld, Multivariate Zeta Function
• Wikipedia, Multiple zeta function

Crossrefs

Cf. A072691, A197110, A258947.

Programs

Mathematica

digits = 99; zetamult[6, 2] = NSum[HarmonicNumber[m-1, 2]/m^6, {m, 2, Infinity}, WorkingPrecision -> digits+20, NSumTerms -> 200, Method -> {"NIntegrate", "MaxRecursion" -> 18}]; zetamult[5, 3] = 5*Zeta[3]*Zeta[5] - (147/24)*Zeta[8] - (5/2)*zetamult[6, 2]; Join[{0}, RealDigits[zetamult[5, 3], 10, digits] // First]

PARI

zetamult([5,3]) \\ Charles R Greathouse IV, Jan 21 2016

Formula

zetamult(5,3) = Sum_{m>=2} (sum_{n=1..m-1} 1/(m^5*n^3)).

Equals Sum_{m>=2} (H(m-1, 3)+polygamma(2,1)/2+zeta(3))/m^5, where H(n,3) is the n-th harmonic number of order 3.

Also equals Sum_{m>=2} (polygamma(2,m)+zeta(3))/(2m^5).

Also equals 5*zeta(3)*zeta(5) - (147/24)*zeta(8) - (5/2)*zetamult(6, 2), where zetamult(6,2) is A258947.