求1+1/3+1/5+1/7+....+1/(2*n-1)

以下是求1+1/3+1/5+1/7+....+1/(2*n-1)的方法：

#include <stdio.h>

float sum(int n);

int main() {
    int n;
    printf("请输入n的值：");
    scanf("%d", &n);
    printf("1+1/3+1/5+1/7+....+1/(2*n-1)=%f\n", sum(n));
    return 0;
}

float sum(int n) {
    float s = 0;
    int i;
    for (i = 1; i <= n; i++) {
        s += 1.0 / (2 * i - 1);
    }
    return s;
}