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Find geometric sum of the series using recursion

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  • Difficulty Level : Easy
  • Last Updated : 09 Aug, 2021

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)


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