java - 编写一个程序，计算 (a/1)+(a/2)+(a/3)+(a/4)+........+(a/n) 之和

Question

我正在尝试解决这个程序，打印其总和:(a/1)+(a/2)+(a/3)+(a/4)+........ ..+(a/n) 其中 a 由用户输入，n 的限制也由用户输入，这是我尝试的程序:

/**
 * Program to 
 * 
 * Anirudh Gupta
 * th August 2014
 */
import java.io.*;
public class Program87b
{
   public static void main () throws IOException
   {
       InputStreamReader isr=new InputStreamReader(System.in);
       BufferedReader br= new BufferedReader(isr);
       System.out.println("Enter the value of the numerator");
       double a=Double.parseDouble(br.readLine());
       System.out.println("Enter the limit for the denominator");
       int limit=Integer.parseInt(br.readLine());
       double n=1.0;
       double sum=0.0;
       while(n<=limit)
       {
           sum=+(a/n);
           n++;
       }
       System.out.println(sum);
   }
}

但是当我输入 a=4 和 n=5 时，我得到 0.8，这只是 (4/5) 的 ans，而不是 (4/1)+(4/2)+(4/3) 的总和+(4/4)+(4/5) 应该是 9.1333333333...

Answer 1

将 =+ 更改为 +=，我不确定变量“d”来自哪里，但我认为应该将其替换为“n”

public static void main () throws IOException
   {
       InputStreamReader isr=new InputStreamReader(System.in);
       BufferedReader br= new BufferedReader(isr);

       System.out.println("Enter the value of the numerator");
       double a=Double.parseDouble(br.readLine());

       System.out.println("Enter the limit for the denominator");
       int limit=Integer.parseInt(br.readLine());

       double n=1.0;
       double sum=0.0;

       while(n<=limit)
       {
           sum+=(a/n);
           n++;
       }

       System.out.println(sum);
   }