Sum of series 8/10, 8/100, 8/1000, 8/10000. . . till N terms

Given a positive integer n, the task is to find the sum of series 

8/10 + 8/100 + 8/1000 + 8/10000. . . till Nth term

Examples:

Input: n = 3
Output: 0.888

Input: n = 5
Output: 0.88888

Approach:

The total sum till nth term of the given G.P. series can be generalized as-

The above formula can be derived following the series of steps-

The given G.P. series

Here,

Thus, using the sum of G.P. formula for r<1

Substituting the values of a and r in the above equation

Illustration:

Input: n = 3
Output: 0.888
Explanation:

S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{3})
        = 0.888 * 0.999
        = 0.888

Below is the implementation of the above problem-

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate sum of
// given series till Nth term
double sumOfSeries(double N)
{
    return (8 * ((pow(10, N) - 1) / pow(10, N))) / 9;
}

// Driver code
int main()
{
    double N = 5;
    cout << sumOfSeries(N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
public class GFG
{

    // Function to calculate sum of
    // given series till Nth term
    static double sumOfSeries(double N)
    {
        return (8
                * ((Math.pow(10, N) - 1) / Math.pow(10, N)))
            / 9;
    }

    // Driver code
    public static void main(String args[])
    {
        double N = 5;
        System.out.print(sumOfSeries(N));
    }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach

# Function to calculate sum of
# given series till Nth term
def sumOfSeries(N):
    return (8 * (((10 ** N) - 1) / (10 ** N))) / 9;

# Driver code
N = 5;
print(sumOfSeries(N));

# This code is contributed by gfgking

C#

// C# program to implement
// the above approach
using System;
class GFG
{

    // Function to calculate sum of
    // given series till Nth term
    static double sumOfSeries(double N)
    {
        return (8
                * ((Math.Pow(10, N) - 1) / Math.Pow(10, N)))
            / 9;
    }

    // Driver code
    public static void Main()
    {
        double N = 5;
        Console.WriteLine(sumOfSeries(N));
    }
}

// This code is contributed by ukasp.

Javascript

<script>
        // JavaScript code for the above approach

        // Function to calculate sum of
        // given series till Nth term
        function sumOfSeries(N) {
            return (8 * ((Math.pow(10, N) - 1) / Math.pow(10, N))) / 9;
        }

        // Driver code
        let N = 5;
        document.write(sumOfSeries(N));

  // This code is contributed by Potta Lokesh
    </script>

Output

0.88888

Time Complexity: O(log n)
Auxiliary Space: O(1)