Help with loop

[lifelesschild]

The question is:
You can approximate e by using the following series
1+1/1!+1/2!+1/3!+1/4!+⋯
Write a program that finds out how many TERMS of this series you need before you first get 2.71828183 approximately.

I am having troubles writing my actual program. I am trying to loop and test then count until i hit the number but i feel as if my program is not what the question is asking. I get that the question is asking how many terms are needed to find the approximation of e and not the number given because the actual approximation is 2.71828182845904523, which is precise.

Below is my program. I would really appreciate some feedback.

public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		double sum=1;
		double k=1;
		while (k<=10)
		{
			sum=sum+(1/factorial (k));
			k++;
			if (sum>=2.71828183)
				break;
		}
		double term=k-1;
	JOptionPane.showMessageDialog (null,"You need "+term+" terms before you first get 2.71828183 approximately. ");
	}
	
	public static double
	factorial (double k) {
		double fact=1;
		for (double j=1;j<=k;j++)
			fact=fact*j;
		return fact;
	}

}

[pbrockway2]

Assuming that to "first get 2.71828183" means that you first obtain a value between 2.718281825 and 2.718281835 then that is what you should test for. (With a bit of thinking you might be able to simplify the test: but those are the values between which numbers round to 2.71828183.)

[lifelesschild]

how would you obtain the value between 2.718281825 and 2.718281835?

[lifelesschild]

when i compile and run it, it says 10 terms are needed

[pbrockway2]

how would you obtain the value between 2.718281825 and 2.718281835?

I'm not sure I understand this. You obtain the value by adding the terms of the series together. The point is to understand when (after how many terms) you reach a value within these limits.

[berb12]

Your while loop stops at when k=10, don't you want it to stop when (sum<e)?

[berb12]

or (sum>=e) for that matter