Java program to find sum of harmonic series 1 + 1/2 + 1/3 + 1/4 + 1/5 +……+ 1/n

Java program to find sum of harmonic series 1 + 1/2 + 1/3 + 1/4 + 1/5 +......+ 1/n

class Harmonic
{
public static void main(String…s)
{
int n,i;
float sum=0;

n=Integer.parseInt(s[0]);

for(i=1;i<=n;i++)
{
sum=sum+(float)1/i;
}

System.out.println(“nSum=”+sum);
}
}

16 thoughts on “Java program to find sum of harmonic series 1 + 1/2 + 1/3 + 1/4 + 1/5 +……+ 1/n

  1. deepa

    class Harmonic
    {
    public static void main(String…s)
    {
    int n,i;
    float sum=0;

    n=Integer.parseInt(s[0]);

    for(i=1;i<=n;i++)
    {
    sum=sum+(float)1/i;
    }

    System.out.println(“nSum=”+sum);
    }
    }

    Reply
  2. Jatin Singh

    Plz tell the code for harmonic series
    S = 1-1/2 + 1/3 – 1/4…… -1/10

    Reply
    1. Ayush tripathi

      class Harmonic
      {
      public static void main(String…s)
      {
      int n,i;
      float sum=0;

      n=Integer.parseInt(s[0]);
      double sign=1;
      for(i=1;i<=n;i++)
      {
      sum=sum+((float)1/i*sign);
      sign*=-1;
      }

      System.out.println(“nSum=”+sum);
      }
      }

      Reply
  3. sriram krishna

    I want Harmonic series program for the below expression.

    1/(n+1)+1/(n+2)+1/(n+3)+………1/(n+n);
    please,help me for this.

    Reply
    1. sriram krishna

      still there no update for my request.please send the reply for the request

      Reply
  4. Samreen fatma

    How to find the sum of series
    Question:
    1*1+ 2*2+3*3*3+4*4+5*5+6*6*6+……

    Reply
  5. Deepanshu Vashist

    why is (float) used in following statement?
    sum=sum+(float)1/i;

    Reply
    1. user

      Because if you divide 1 by i which is an integer your ans will be an integer. E.g. if i=2, 1/i will give you 0 and not 0.5, that’s why you typecast either numerator or denominator to float, so that the division 1/i is also float

      Reply
      1. soorajmahato

        import java.util.Scanner;
        class sum_of_hp
        {
        public static void main(String args[])
        {
        double num,i,sum=0;;
        Scanner sc=new Scanner(System.in);
        System.out.print(“upto how many terms you want to sum = “);
        num=sc.nextInt();
        for(i=1;i<=num;i++)
        {
        sum=sum+(1/i);
        }
        System.out.println("sum of the series upto "+num+" terms is " +sum);
        }
        }
        instead of using typecasting why didnot you take double type it is easier than type casting

        Reply
      2. sooraj

        import java.util.Scanner;
        class sumofhp
        {
        public static void main(String args[])
        {
        double num,i;
        double sum=0;
        Scanner sc=new Scanner(System.in);
        System.out.print(“upto how many terms you want to sum = “);
        num=sc.nextInt();
        for(i=1;i<=num;i++)
        {
        sum=sum+(1/i);
        }
        System.out.println("sum of the series upto "+num+" terms is " +sum);
        }
        }

        Reply
  6. sooraj

    instead of taking two different data type int and float can i use double type?

    Reply
  7. Fatima

    import java.util.Scanner;

    public class sumofseries {
    public static void main(String[] args)
    {
    double sum = 0;
    int n;
    System.out.println(“1!/1+2!/2+3!/3+4!/4+5!/5”);
    Scanner s = new Scanner(System.in);
    System.out.print(“Enter the no. of terms in series:”);
    n = s.nextInt();
    sumofseries obj = new sumofseries();
    for(int i = 1; i 0)
    {
    mul = mul * x;
    x–;
    }
    return mul;
    }
    }

    Reply
  8. Kishan Soni

    How to find 9th n for the below formula
    H(n,r) = 1/1^r + 1/2^r + … + 1/n^r

    Reply

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