Geometric sum using recursion in python

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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