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0.0001import 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");  }   }}