编程 设s=1+1/2+1/3+...+1/n,求与8最接近的的s的值及与之对应的n值

//marhNum.cpp
#include<iostream>
using namespace std;
double addNum(int num);
int main()
{
double s=1;
int n=3;
double maxNum=0;
while (s<8){
double num=0;
maxNum=s;
num=addNum(n);
s=s num;
n ;
};
cout<<"s="<<maxNum<<endl;
cout<<"n="<<n-1;
return 0;
}
double addNum(int num)
{
double s=0;
s=(double)1/num;
return s;
}
----------------------------
S:7.99966
n:2759