The given series 0.6,0 .o6,.... is a geometric progression where each element is the previous element divided by 10. So find the sum of the series we have to apply the sum of GP one formula for r less than 1(r=0.1 in our case).
Sum = 6/10 [1- (1/10)n/(1-1/10)] Sum = 6/9 [1- (1/10)n] Sum = 2/3[1- (1/10n)]
Example
#include <stdio.h>
#include <math.h>
int main() {
int n = 6;
float sum = 2*((1 - 1 / pow(10, n)))/3;
printf("sum = %f", sum);
}Output
sum = 0.666666