# 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. Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

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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 using namespace std; // Function to find the sum of seriesvoid 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 Codeint main(){    int N = 10;     printSeriesSum(N);    return 0;}

## Java

 // Java program for the above approachclass GFG{  // Function to find the sum of seriesstatic 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 Codepublic 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 seriesdef 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 Codeif __name__ == "__main__" :     N = 10;     printSeriesSum(N);        # This code is contributed by Yash_R

## C#

 // C# program for the above approachusing System;  class GFG {      // Function to find the sum of seriesstatic 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

 
Output:
-0.263456

Time Complexity: O(N)

Auxiliary Space: O(1)

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