Find the Sum of the series 1/2 - 2/3 + 3/4 - 4/5 + ... till N terms

Given a number N, the task is to find the sum of the below series till N terms.

\frac{1}{2} - \frac{2}{3} + \frac{3}{4} - \frac{4}{5} + ... 

Examples:

Input: N = 6 
Output: -0.240476
Input: N = 10 
Output: -0.263456 

Approach: From the given series, find the formula for Nth term: 

1st term = 1/2
2nd term = - 2/3
3rd term = 3/4
4th term = - 4/5
.
.
Nthe term = ((-1)N) * (N / (N + 1))

Therefore: 

Nth term of the series 

Then iterate over numbers in the range [1, N] to find all the terms using the above formula and compute their sum.
Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to find the sum of series
void printSeriesSum(int N)
{
    double sum = 0;

    for (int i = 1; i <= N; i++) {

        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if (i & 1) {
            sum += (double)i / (i + 1);
        }
        else {
            sum -= (double)i / (i + 1);
        }
    }

    // Print the sum
    cout << sum << endl;
}

// Driver Code
int main()
{
    int N = 10;

    printSeriesSum(N);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to find the sum of series
static void printSeriesSum(int N)
{
    double sum = 0;
 
    for (int i = 1; i <= N; i++) {
 
        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if (i % 2 == 1) {
            sum += (double)i / (i + 1);
        }
        else {
            sum -= (double)i / (i + 1);
        }
    }
 
    // Print the sum
    System.out.print(sum +"\n");
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 10;
 
    printSeriesSum(N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to find the sum of series
def printSeriesSum(N) :
    
    sum = 0;

    for i in range(1, N + 1) :

        # Generate the ith term and
        # add it to the sum if i is
        # even and subtract if i is
        # odd
        if (i & 1) :
            sum += i / (i + 1);
     
        else :
            sum -= i / (i + 1);
    

    # Print the sum
    print(sum);

# Driver Code
if __name__ == "__main__" :

    N = 10;

    printSeriesSum(N);
   
    # This code is contributed by Yash_R

C#

// C# program for the above approach
using System;
 
class GFG {
     
// Function to find the sum of series
static void printSeriesSum(int N)
{
    double sum = 0;

    for (int i = 1; i <= N; i++) {

        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if ((i & 1)==0) {
            sum += (double)i / (i + 1);
        }
        else {
            sum -= (double)i / (i + 1);
        }
    }

    // Print the sum
    Console.WriteLine(sum);
}

// Driver Code
    public static void Main (string[] args)
    {
       
    int N = 10;

    printSeriesSum(N);
}
}

// This code is contributed by shivanisinghss2110

JavaScript

<script>

// javascript program for the above approach

// Function to find the sum of series
function printSeriesSum( N)
{
    let sum = 0;

    for (let i = 1; i <= N; i++) {

        // Generate the ith term and
        // add it to the sum if i is
        // even and subtract if i is
        // odd
        if (i & 1) {
            sum += i / (i + 1);
        }
        else {
            sum -= i / (i + 1);
        }
    }

    // Print the sum
     document.write( sum.toFixed(6) );
}

// Driver Code

    let N = 10;

    printSeriesSum(N);
   
// This code is contributed by todaysgaurav 

</script>

Output:

-0.263456

Time complexity: O(n) for given input n

Auxiliary Space: O(1), since no extra space has been taken.

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