由命令行输入一个正整数n，求下列式子的和。
(a)s=1+1/2+1/3+1/4+...+1/n
(b)s=1-1/2+1/3-1/4+...+1/n
(c)s=1+1/(1*2)+1/(2*3)+...+1/(n*(n+1))
(d)s=1/1!+1/2!+1/3!+...1/n!
/*由命令行输入一个正整数n，求下列式子的和。
(a)s=1+1/2+1/3+1/4+...+1/n
(b)s=1-1/2+1/3-1/4+...+1/n
(c)s=1+1/(1*2)+1/(2*3)+...+1/(n*(n+1))
(d)s=1/1!+1/2!+1/3!+...1/n!
*/
public class Xiti4q16 {
public static void main(String args[]) {
// a
int n = Integer.parseInt(args[0]);
double sum_a = 0;
for (double i = 1; i <= n; i++)
sum_a = sum_a + (1 / i);
System.out.println("(a)s=1+1/2+1/3+1/4+...+1/n的值是：" + sum_a);
// b
double sum_b = 0;
int j = 1;
for (double i = 1; i <= n; i++, j = -j) {
sum_b = sum_b + (1 / (i * j));
}
System.out.println("(b)s=1-1/2+1/3-1/4+...+1/n的值是：" + sum_b);
// c
double sum_c = 0;
for (double i = 1; i <= n; i++) {
sum_c = sum_c + (1 / (i * (i + 1)));
}
System.out
.println("(c)s=1+1/(1*2)+1/(2*3)+...+1/(n*(n+1))的值是：" + sum_c);
//d
double sum_d=0;
for (double i=1; i<=n; i++){
double sum_n=1;
for (double k=1;k<=i;k++){
sum_n=sum_n*k;
// System.out.println(k +"的阶乘是：" + sum_n);
}
sum_d=sum_d+(1/sum_n);
}
System.out.println("(d)s=1/1!+1/2!+1/3!+...1/n! 的值是：" + sum_d);
}
}
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