数列求和

1+1/2+1/3+…+1/n 从右往左更加精确

import java.util.Scanner;
class Main {
  public static void main(String[] args){
    int n = 50_000;
    double sum = 0;
    for(int i = 1; i <= n; i++){//从左往右加
      sum += 1.0 / i;
    }
    System.out.print("The result is: " + sum);

    double sum1 = 0;
    for(int i = n; i > 0; i--){//从右往左加 更加精确
      sum1 += 1.0 / i;
    }
    System.out.print("\nThe result is: " + sum1);
  }
}

1/3 + 3/5 + 5/7+…+ 95/97 + 97/99

import java.util.Scanner;
class Main {
  public static void main(String[] args){
    double sum = 0;
    for(int i = 1; i <= 97;i = i+2)
      sum += (double)i / (i + 2);

    System.out.print("The sum is: " + sum);
    
  }
}

PI = 4 * (1-1/3+1/5-1/7+…+1/(2i-1)-1/(2i+1))

显示当i为10,000 20,000 … 100,000时PI的值

import java.util.Scanner;
class Main {
  public static void main(String[] args){
    double PI = 0;
    double result = 0;
    int n = 10_000;
    System.out.print("i\t\tPI");
    while(n <= 100_000){
      for(int i = 1; i <= n; i = i + 2)
        result += 1.0 / (2 * i -1) - 1.0 / (2 * i + 1);
      
      PI = 4 * result;
      System.out.print("\n" + n + "\t" + PI);
      result = 0;
      n += 10_000;
    }
  }
}

e = 1 + 1/1! + 1/2! +…+1/i!

当i为10,000 20,000 … 100,000时e的值

import java.util.Scanner;
class Main {
  public static void main(String[] args){
    
    System.out.print("i\t\te");
    for(int n = 10_000;n <= 100_000; n += 10_000){
      double e = 0;
      
      for(int i = 1; i <= n; i++){
        double iFactorial = 1.0; //i的阶乘的倒数
        for(int j = i; j > 0; j--)
          iFactorial = iFactorial * 1.0 / j;
        e += iFactorial; 
      }
      e += 1.0;
      System.out.print("\n" + n + "\t" + e);
    }      
  }
}

求下面的数列的值

$$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{624}+\sqrt{625}}$$

import java.util.Scanner;
class Main {
  public static void main(String[] args){
    int n = 624;
    double sum = 0;
    for(int i = 1; i <= n; i++){
      sum += 1.0 / (i + Math.pow((i + 1), 0.5));
    }
    System.out.print("The result is: " + sum);
  }
}