Sum of series 2/3 – 4/5 + 6/7 – 8/9 + ...... upto n terms

In C programming, we can calculate the sum of the series 2/3 − 4/5 + 6/7 − 8/9 + ...... up to n terms. This is an alternating series where each term has the form (-1)n+1 * (2*n) / (2*n + 1).

Syntax

double calculateSeriesSum(int n);

Understanding the Series Pattern

The series follows this pattern −

• First term: 2/3 (positive)
• Second term: -4/5 (negative)
• Third term: 6/7 (positive)
• Fourth term: -8/9 (negative)

For the nth term: (-1)n+1 * (2*n) / (2*n + 1)

Example: Calculate Series Sum

Here's a complete C program to calculate the sum of this series −

#include <stdio.h>

double calculateSeriesSum(int n) {
    double sum = 0.0;
    int sign = 1;
    
    for (int i = 1; i <= n; i++) {
        double term = (double)(2 * i) / (2 * i + 1);
        sum += sign * term;
        sign *= -1;  // Alternate the sign
    }
    
    return sum;
}

int main() {
    int n1 = 10, n2 = 17;
    
    printf("For n = %d:<br>", n1);
    printf("Sum = %.6f<br><br>", calculateSeriesSum(n1));
    
    printf("For n = %d:<br>", n2);
    printf("Sum = %.6f<br>", calculateSeriesSum(n2));
    
    return 0;
}

For n = 10:
Sum = -0.191921

For n = 17:
Sum = 0.071520

How It Works

• We use a loop to generate each term of the series
• For each iteration i, the numerator is 2*i and denominator is 2*i + 1
• We alternate the sign using a variable that switches between 1 and -1
• Each term is added to the running sum

Alternative Implementation

Here's another approach using the mathematical formula directly −

#include <stdio.h>
#include <math.h>

int main() {
    int n = 10;
    double sum = 0.0;
    
    printf("Series: ");
    for (int i = 1; i <= n; i++) {
        double term = pow(-1, i + 1) * (2.0 * i) / (2 * i + 1);
        sum += term;
        
        if (i == 1) {
            printf("%.0f/%d", 2.0 * i, 2 * i + 1);
        } else {
            printf(" %c %.0f/%d", (term > 0) ? '+' : '-', 
                   fabs(2.0 * i), 2 * i + 1);
        }
    }
    
    printf("\nSum = %.6f<br>", sum);
    return 0;
}

Series: 2/3 - 4/5 + 6/7 - 8/9 + 10/11 - 12/13 + 14/15 - 16/17 + 18/19 - 20/21
Sum = -0.191921

Key Points

• The series alternates between positive and negative terms
• Each term follows the pattern (2*n)/(2*n + 1)
• Use double data type for accurate floating-point calculations
• The sign can be controlled using pow(-1, i+1) or a toggle variable

Conclusion

This alternating series can be efficiently calculated using a simple loop that generates each term and maintains the alternating sign pattern. The sum converges to different values depending on the number of terms.