(1)黎曼zeta函数表达式
ζ(s)=1/1s+1/2s+1/3s+...+1/ms (m趋于无穷,且m始终是偶数)
(2)将表达式两边同时乘以(1/2s)
(1/2s)*ζ(s)=1/1s*1/2s+1/2s*1/2s+1/3s*1/2s+...+1/ms*1/2s=1/2s+1/4s+1/6s+...+1/(2*m)s
由(1)-(2)得
ζ(s)-(1/2s)*ζ(s)=1/1s+1/2s+1/3s+...+1/ms-[1/2s+1/4s+1/6s+...+1/(2*m)s]
而在欧拉乘积公式推导结果如下
ζ(s)-(1/2s)*ζ(s)=1/1s+1/3s+1/5s+...+1/(m-1)s.