Compute and approximat value for euler's constant

[Mcody2]

the problem is for me to compute and approxiamte value for euler's constant by summing the series (adding up the terms 1, 1/2, ...., 1/n in that order and then subtracting the value fo In(n)).
I have to do this with both double and float arithmetic.

Here is what I have so far.

import java.io.*;
public class PP2 {
public static void main(String[] args) throws

IOException {
System.out.println("Enter the limit for sum:");
BufferedReader kbd = new BufferedReader(new

InputStreamReader(System.in));
String S = kbd.readLine();
int loops = Integer.parseInt(S.trim());
int n;
for (n = 1; n <= loops; n++)
{
double sum = sum + (double)(1/n);
float sum2 = sum2 + (float)(1/n);
}
double condoub = sum - (double)Math.log(loops);
float conflout = sum2 - (float)Math.log(loops);
System.out.println("Euler's const, double: =" + condoub);
System.out.println("Euler's const, flout: =" + conflout);
}
}
thanks for the help

[sjalle]

You should declare the sum and sum2 prior to the loop, the way you have
coded it will not accumulate, it doesn't even compile

import java.io.*;

public class PP2 {
  public static void main(String[] args) throws IOException {

    System.out.println("Enter the limit for sum:");
    BufferedReader kbd = new BufferedReader(new InputStreamReader(System.in));
    String S = kbd.readLine();
    int loops = Integer.parseInt(S.trim());
    int n;
    double sum=0;
    float sum2=0;
    for (n = 1; n <= loops; n++) {
      sum = sum + (double) (1 / n);
      sum2 = sum2 + (float) (1 / n);
    }
    double condoub = sum - (double) Math.log(loops);
    float conflout = sum2 - (float) Math.log(loops);
    System.out.println("Euler's const, double: =" + condoub);
    System.out.println("Euler's const, flout: =" + conflout);
  }
}

[Mcody2]

Thanks I just fixed that it works fine now.