编程计算， sn=(1-1/2)+(1/3-1/4)+...+(1/(2n-1)-1/(2n))，结果保留两位小数。c语言

#include <stdio.h>

int main() {
int n;
double sn = 1;
printf("请输入 n 的值：");
scanf("%d", &n);
for (int i = 2; i <= 2 * n; i += 2) {
sn *= (double)(i - 1) / i;
}
printf("结果为：%.2lf", sn);
return 0;
}