public class TestCode {

    public static double getSun(double in) {
        double sun = 0;
        for (double i = 2; i <= in; i += 2) {
            sun = sun + 1 / i;
        }
        return sun;
    }

    public static void main(String[] args) {
        System.out.println(getSun(40));
    }
}

------解决方案--------------------
探讨

Java code

public class TestCode {

    public static double getSun(double in) {
        double sun = 0;
        for (double i = 2; i <= in; i += 2) {
            sun = sun + 1 / i;
        }
       ……


------解决方案--------------------
探讨

非常感谢，其实我想要的是带分子和分母的结果


------解决方案--------------------
Java code
public class Test {
    public static void main(String[] args) throws Throwable {
        
        int min = 2;
        for (int i=4; i<=40; i+=2) { //求分母的最小公倍数
            min = getMinTimes(min, i);
        }
        
        int sum = 0;
        for (int i=2; i<=40; i+=2) { //通分分子求和, 上面这里写错了，是i+=2，不是每次递增1
            sum += min/i;
        }

        System.out.printf("%d/%d=%.4f\n", sum, min, ((double)sum)/min); //输出结果

        double n = 0;
        for (int i=2; i<=40; i+=2) { //检验结果，上面这里写错了
            n += ((double)1)/i;
        }
        System.out.printf("%.4f\n", n);
    }

    public static int getMinTimes(int a, int b) { //求两个数的最小公倍数
        int a1 = a, b1 = b;
        int mod = a%b;
        while (mod != 0) {
            a = b;
            b = mod;
            mod = a%b;
        }
        return (a1*b1)/b;
    }
}

------解决方案--------------------
探讨

+1
母子分母只差约分了.


------解决方案--------------------
for example :

Java code
//题目：1/2+1/4+1/6+1/8+.........+1/40的和，求java算法，

public class Test {
    public static void main(String[] args) {
          float sum = 0 ;
          //int i ;

          for ( float i = 2 ; i <= 40 ; i += 2 )
              sum += 1 / i ;

          System.out.println ("1/2+1/4+1/6+1/8+.........+1/40 = " + sum );
    }
}

------解决方案--------------------
Java code


public class B {
    public static void main(String[] args) {
        long[] result = new long[]{1,2};
        for(int i = 1 ; i < 20; i++){
            result = sum(1, i*2 + 2 , result);
        } 
        System.out.println(result[0] + "/" + result[1]);
    }
    private static long[] sum(long fz,long fm, long result[]){
        long[] result1 = new long[2];
        result1[0] = result[0] * fm + fz * result[1];
        result1[1] = result[1] * fm;
        long max = 1;
        for(long i = 1 ; i <= result1[0]; i++){
            if(result1[0]%i == 0 && result1[1]%i ==0){
                max = i;
            }
        } 
        if(max != 1){
            result1[0] = result1[0]/max;
            result1[1] = result1[1]/max;
        }  
        System.out.println(result1[0] + "/" + result1[1]);
        return result1;
    }

}

------解决方案--------------------