Geometric sum using recursion in python

View Discussion

Improve Article

Save Article

View Discussion

Improve Article

Save Article

Given an integer N, we need to find the geometric sum of the following series using recursion. 
 

1 + 1/3 + 1/9 + 1/27 + … + 1/(3^n) 
 

Examples: 
 

Input N = 5 
Output: 1.49794

Input: N = 7
Output: 1.49977

Approach:
In the above-mentioned problem, we are asked to use recursion. We will calculate the last term and call recursion on the remaining n-1 terms each time. The final sum returned is the result.
Below is the implementation of the above approach: 
 

C++

#include

using namespace std;

double sum(int n)

{

    if (n == 0)

        return 1;

    double ans = 1 / (double)pow(3, n) + sum(n - 1);

    return ans;

}

int main()

{

    int n = 5;

    cout << sum(n) << endl;

    return 0;

}

Java

import java.util.*;

class GFG {

    static double sum(int n)

    {

        if (n == 0)

            return 1;

        double ans = 1 / (double)Math.pow(3, n) + sum(n - 1);

        return ans;

    }

    public static void main(String[] args)

    {

        int n = 5;

        System.out.println(sum(n));

    }

}

Python3

def sum(n):

    if n == 0:

        return 1

    return 1 / pow(3, n) + sum(n-1)

n = 5;

print(sum(n));

C#

using System;

class GFG {

    static double sum(int n)

    {

        if (n == 0)

            return 1;

        double ans = 1 / (double)Math.Pow(3, n) + sum(n - 1);

        return ans;

    }

    static public void Main()

    {

        int n = 5;

        Console.WriteLine(sum(n));

    }

}

Javascript

Time Complexity: O(N)
Auxiliary Space: O(N)