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

• Difficulty Level : Expert
• Last Updated : 16 Aug, 2022

Given a number N, the task is to find the sum of the below series till N terms. 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) for given input n

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

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