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,要求誤差小於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("在文本區輸入數字n後回車",10,100); g.drawString("文本區顯示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("請輸入數字字符"); } } else if(e.getSource()==Button1) { text1.setText("0"); } }}