eps precision summation and its own problem analysis

//Add to the first item that does not meet the conditions. Not the last item that meets the criteria.
//The first item that does not meet the conditions is adjacent to the last item that meets the conditions.
method one

#include<stdio.h>
#include<math.h>
void main()
{
    
    
    int i;
    double sum=0,eps;
    scanf("%lf",&eps);
    for(i=1;1*1.0/(3*i-2)>=eps;i++)//这个for循环判断的是当前项是否符合条件。所以那个不符合条件的一项也没加
    {
    
    
        sum+=pow((-1),(i+1))*1.0/(3*i-2);
    }
	sum+=pow((-1),(i+1))*1.0/(3*i-2);//加上了
    printf("sum = %.6lf",sum);
}

Method Two

#include<stdio.h>
#include<math.h>
void main(){
    
    
	double sum=0,i=1,esp;
	scanf("%lf",&esp);
	do{
    
    
		sum+=pow(-1,(i+1))*1.0/(3*i-2);
		i++;
	}while(1.0/(3*(i-1)-2)>esp);//这样就是先加上再判断当前项是否符合。当不符合条件的第一项加上后，再判断跳出循环
    printf("sum = %.6lf",sum);
}

Own error analysis

#include<stdio.h>
#include<math.h>
int main(){
    
    
	int flag=1, i=1;//符号和分母定义 
	double sum=0,n=1.0,eps; //初始复制很关键
	scanf("%lf",&eps);
	if(eps>=1) //判断当eps大于等于1时临界情况
		printf("sum = %.6f",n);
	else
	{
    
    
		while(fabs(n)>eps)//判断的是上一项的不是下一项的，while里第一遍第二遍判断的都是第一项1！！
		{
    
    
			n=flag*1.0/i;
			sum=sum+n;
			i+=3;
			flag=-flag;
		}
		printf("sum = %.6f",sum);
	}
	return 0;
}