C Primer Plus6-11

/*11.考虑这两个无限序列：1.0+1.0/2.0+1.0/3.0+1.0/4.0+…1.0 - 1.0/2.0+1.0/3.0 - 1.0/4.0+…编写一个程序来计算这两个序列不断变化的总和，直到达到某个次数。让用户交互地输入这个次数。看看在 20 次、100 次和 500 次之后的总和。是否每个序列都看上去要收敛于某个值？提示：奇数个-1相乘的值为-1，而偶数个-1 相乘的值为 1。*/#include<stdlib.h>#include<stdio.h>int main(void){  double i, sum1, sum2;  int sign, number;    printf("Enter the number:");    scanf("%d",&number); for (i = 1.0, sign = 1, sum1 = 0, sum2 = 0; i <= number; i++, sign *= -1)  {    sum1 = sum1 + 1.0 / i;    sum2 = sum2 + sign * 1.0 / i;  }    printf("1.0 + 1.0/2.0 + 1.0/3.0 + 1.0/4.0 ...(%d item) = %lf\n", number, sum1);    printf("1.0 - 1.0/2.0 + 1.0/3.0 - 1.0/4.0 ...(%d item) = %lf\n", number, sum2);    system("pause");    return(0);}