计算1-2/3!+3/5!-4/7!+5/9!-6/11!+……的和，直到某一项的绝对值小于10 的-6次方为止

#include <stdio.h>
int main()
{
    double item = 1,sum=1,fm=1;
    int fz = 1;
    int m = 1;
    int flag = 1;
    do
    {
        flag = -flag;
        fz++;
        fm *= (m+1)*(m+2);
        m+=2;
        item = fz*1.0/fm;
        sum += item * flag;
    }while(item >= 1e-6);
    printf("%lf",sum*4);
}