ζ(k)的部分和五阶和式的计算  

Computing the 5-Order Sums of ζ(k)

作　　者：孙平[1] 

机构地区：[1]东北大学数学系,沈阳110006

出　　处：《数学学报（中文版）》2003年第2期297-302,共6页Acta Mathematica Sinica：Chinese Series

摘　　要：u1,u2…是独立、同分布于(0,1)区间上均匀分布的随机变量.本文证明了1-u1u2…uk的n-1阶矩(n≥1)是以调和数的部分和ζn(r)=∑j=1n 1/jr,r≥1为变元的指数型完全Bell多项式,因此Riemann-Zeta函数ζ(k),k≥2能够被展开成第一类无符号Stirling数s(n,k)的级数,从而计算出与ζn(r)有关的全部6个五阶和式.它们都是ζ(5)与ζ(2)ζ(3)的有理组合.Let u1, u2, … be independent random variables uniformly distributed on (0,1). This paper derives that for any n ≥ 1, the (n - 1)th moments of 1 - u1u2… uk are exponential complete Bell polynomials Yk(ζn(1), 1!ζn(2), 2!ζn(3), 3!ζn(4),…). Here the partial sums ζn(r) =∑j=1n/jr, r≥1, so the Riemann-Zeta function ζ(k) can be expanded as the series involving Stirling numbers of the first kind. Furthermore, all the 5-order sums of ζn(r) such as ∑n>1 Hn2/n3 have been given.

关 键 词：Riemann-Zeta函数 STIRLING数 BELL多项式 均匀分布 矩 

分 类 号：O156.4[理学—数学]