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Find the Sum of the series 1/2 – 2/3 + 3/4 – 4/5 + … till N terms

  • Last Updated : 24 Nov, 2021

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} + ...

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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 

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

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)

Auxiliary Space: O(1)




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