求出e=1+1/1!+1/2!+1/3!+……+1/n!+……的近似值的java applet程序

//求出e=1+1/1!+1/2!+1/3!+……+1/n!+……的近似值,要求误差小于0.0001
import java.applet.*;
import java.awt.*;
import java.awt.event.*;
public class AT1_1 extends Applet implements ActionListener
{ 
 TextField text1;
 Button Button1;
 
 public void init()
 {
  text1 = new TextField("0",10);
  Button1 = new Button("清除");
  add(text1);
  add(Button1);
  text1.addActionListener(this);
  Button1.addActionListener(this);
 }
 public void start(){}
 public void stop(){}
 public void destory(){}
 public void paint(Graphics g)
 { 
  g.drawString("在文本区输入数字n后回车",10,100);
  g.drawString("文本区显示1+1/1!+1/2!+1/3!+……+1/n!+……的近似值",10,120);
 }
 public void actionPerformed(ActionEvent e)
 {
  if(e.getSource()==text1)
  {
   double sum=1,a=1;
   int i=1;
   int n=0;
   try
   {
   n = Integer.valueOf(text1.getText()).intValue();
   while(i<=n)
   { 
    a = a*(1.0/i);
    sum = sum + a;
    i=i+1;
   }
   sum=sum*10000;
   sum=Math.round(sum);
   sum=sum/10000;
   text1.setText(""+sum);  
   }
   catch(NumberFormatException Event)
   {
   text1.setText("请输入数字字符");
   }
  }
  else if(e.getSource()==Button1)
  {
   text1.setText("0");
  }
  
 }
}

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