编程计算y=1+1/x+1/(x*x)+1/(x*x*x)+.的值(x>1),直到最后一项小于10的负4次方

#include
double function(int a,double x)
{
if(a>0)
return function(a-1,x)*x;
return 1.000;
}
int main()
{
double i=0,y=0,x;
printf("输入x的值\n");
scanf("%lf",&x);
while((1/function(i,x))>1e-4)
{
y=y+(double)(1/function(i,x));
i++;
}
printf("%lf\n",y);
return 0;
}