Sums of the Numbers of Prime Factors of Positive Integers

Denote by ω(n) and Ω(n) the number of distinct prime factors of n and the total number of prime factors of n,respectively.For any positive integer ι,we prove that ∑↑2≤n≤x1/ω（n)=ι↑∑↑κ=0(ι↑∑↑i=κ（-1)^i-κCi^κF^(i-κ)(1)κ！x/（loglogx）^i+1+O(x/(loglogx)^ι+2) ∑↑2≤n≤xΩ（n)/ω（n)=x+ι↑∑↑κ=0ι↑∑↑i=κ∑↑p1/p^κ+2-p^κ+1（-1)^i-κCi^κF^(i-κ）（1）κ!)x/(loglogx)^i+1+O(x/(loglogx)ι+2） where F(z)=1/г(z)pⅡ（1+z/p-1)(1-1/p)^z,and the constant O despends on ι.This improves previous result of R.L.Duncan and Chao Huizhong.