C Programming for sum of the series 0.6, 0.06, 0.006, 0.0006, ...to n terms

The given series 0.6, 0.06, 0.006, 0.0006, ... is a geometric progression where each term is obtained by dividing the previous term by 10. To find the sum of this series, we apply the formula for the sum of a geometric progression where the common ratio r

Syntax

Sum = a * (1 - r^n) / (1 - r)
where a = first term, r = common ratio, n = number of terms

Mathematical Formula

For the given series −

First term (a) = 0.6 = 6/10
Common ratio (r) = 0.1 = 1/10

Sum = (6/10) * [1 - (1/10)^n] / (1 - 1/10)
Sum = (6/10) * [1 - (1/10)^n] / (9/10)
Sum = (6/9) * [1 - (1/10)^n]
Sum = (2/3) * [1 - (1/10)^n]

Example

Here's a C program to calculate the sum of the series for n terms −

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

int main() {
    int n = 6;
    double sum = (2.0/3.0) * (1 - 1.0/pow(10, n));
    
    printf("Number of terms: %d<br>", n);
    printf("Sum of series: %.6f<br>", sum);
    
    // Display the series terms
    printf("Series: ");
    for(int i = 1; i <= n; i++) {
        printf("%.3f", 6.0/pow(10, i));
        if(i < n) printf(" + ");
    }
    printf(" = %.6f<br>", sum);
    
    return 0;
}

Number of terms: 6
Sum of series: 0.666666
Series: 0.600 + 0.060 + 0.006 + 0.001 + 0.000 + 0.000 = 0.666666

Key Points

• The series is a geometric progression with first term a = 0.6 and common ratio r = 0.1
• The sum approaches 2/3 (? 0.666667) as n approaches infinity
• Each term is exactly 1/10th of the previous term

Conclusion

The sum of the geometric series 0.6 + 0.06 + 0.006 + ... for n terms can be calculated using the GP sum formula. As the number of terms increases, the sum converges to 2/3.