Arithmetic Recursion

Question

I am trying to write a code that computes the following for a given integer n:

1/1 + 1/2 + 1/3 ... + 1/n

This is the code I have written so far:

public class RecursiveSum
{
  public static double Sumto(int n)
  {
    if (n == 0) { return 0.0; }
    else if (n > 0) { return 1/n + 1/Sumto(n - 1); }
    else { throw new IllegalArgumentException("Please provide positive integers"); }
  }
  public static void main(String[] args)
  {
    System.out.println(Sumto(5));
  }
}

However, it always outputs:

Infinity

What is the problem and how can I fix it?

Thank you

Answer 1

You have two issues :

You must perform floating point division (i.e. replace 1/n with 1.0/n), and you should add Sumto(n - 1) to 1.0/n to get Sumto(n).

  public static double Sumto(int n)
  {
    if (n == 0) { return 0.0; }
    else if (n > 0) { return 1.0/n + Sumto(n - 1); }
    else { throw new IllegalArgumentException("Please provide positive integers"); }
  }

The reason you got Infinity was that 1/Sumto(n - 1) returns Infinity when Sumto(n - 1) is 0.0, and Sumto(0) is 0.0.

Answer 2

However, it always outputs: Infinity

Because you are doing 1/0 in the following steps in your code which yields Infinity.

else if (n > 0) { return 1/n + 1/Sumto(n - 1);

You thought n > 0 escapes the n / 0 stuffs, but nope! Think about the case when n = 1 which passes n > 0 case but fall into a trap to:

1/Sumto(n - 1)
1/Sumto(1 - 1)
1/Sumto(0)

where Sumto(0) returns 0.0. Hence,

 1/0.0

yields Infinity. Moreover, use 1.0/n instead of 1/n as it is floating point division.

So add another condition, like

if(n == 1) 
    return 1;