// 2024 IUSB Programming Competition
// Round 1 Problem 1
// Majic Function
// Solution by Liguo Yu

#include <iostream>
#include <cmath>

int main()
{
    int number;
    std::cin >> number;
    double result = 3;
    
    for (int i = 1; i <= number; i++)
    {
        result = result + std::pow(-1, i + 1) * 4.0 / (2 * i * (2 * i + 1) * (2 * i + 2));
    }

    std::cout << result << std::endl;

    return 0;
}