Find geometric sum of the series using recursion

Last Updated : 15 Mar, 2025

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/(3ⁿ)

Examples:

Input: n = 5
Output: 1.49794
Explanation: 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 = 1.49794

Input: n = 7
Output: 1.49977

Approach:

To find the geometric sum of the series 1 + 1/3 + 1/3² + ... + 1/3ⁿ. The base case returns 1 when n = 0. For each recursive call, the function adds 1/3ⁿ to the sum of the remaining terms. The recursion continues until n reaches 0, ensuring all terms are added.

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

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));
    }
}

Python

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));

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

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;
console.log(sum(n).toFixed(5));

Output

1.49794

Time Complexity: O(n)
Auxiliary Space: O(n), due to recursive function calls stored in the call stack.

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