用C语言程序改错：用以下公式计算e的近似值： 精度要求为10-6，计算精度公式为

#include <stdio.h>

int main()
{
int i = 1;
double item = 1.0, sum = 1.0, eps = 1e-6;

while (item > eps)
{
item = item / i;
sum = sum + item;
i++;
}

printf("The approximate value of e is %f", sum);

return 0;
}