Find geometric sum of the series using recursion

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++

// CPP implementation to Find the
// geometric sum of the series using recursion

#include <bits/stdc++.h>
using namespace std;

// function to find the sum of given series
double sum(int n)
{
    // base case
    if (n == 0)
        return 1;

    // calculate the sum each time
    double ans = 1 / (double)pow(3, n) + sum(n - 1);

    // return final answer
    return ans;
}

// Driver code
int main()
{

    // integer initialisation
    int n = 5;

    cout << sum(n) << endl;

    return 0;
}

Java

// JAVA implementation to Find the
// geometric sum of the series using recursion

import java.util.*;

class GFG {

    static double sum(int n)
    {
        // base case
        if (n == 0)
            return 1;

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

        // return final answer
        return ans;
    }

    // Driver code
    public static void main(String[] args)
    {
        // integer initialisation
        int n = 5;

        // print result
        System.out.println(sum(n));
    }
}

Python3

# CPP implementation to Find the
# geometric sum of the series using recursion


def sum(n):

    # base case
    if n == 0:
        return 1

    # calculate the sum each time
    # and return final answer
    return 1 / pow(3, n) + sum(n-1)

n = 5;

print(sum(n));

C#

// C# implementation to Find the
// geometric sum of the series using recursion

using System;

class GFG {

    static double sum(int n)
    {
        // base case
        if (n == 0)
            return 1;

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

        // return final answer
        return ans;
    }

    // Driver code
    static public void Main()
    {
        int n = 5;

        Console.WriteLine(sum(n));
    }
}

Javascript

<script>

// Javascript implementation to Find the
// geometric sum of the series using recursion

// function to find the sum of given series
function sum(n)
{
    // base case
    if (n == 0)
        return 1;

    // calculate the sum each time
    var ans = 1 / Math.pow(3, n) + sum(n - 1);

    // return final answer
    return ans;
}

// Driver code
// integer initialisation
var n = 5;
document.write( sum(n).toFixed(5));

</script>

Output:

1.49794

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