The formal way to solve your exercise:

To check, run the following C program (MinGW2.95 compiler):
#include <stdio.h> #include <math.h> int main() { int sumIN = 0, sumOUT = 0; double i, n = 500, j; double d; for (i = 1; i <= n; i ++) { d = 1/(double)i; j = i; while (j > 0 && d > 0) { j -= d; sumIN ++; } sumOUT ++; } printf("\nsumIN = %d, sumOUT = %d", sumIN, sumOUT); printf("\n"); return 0; }
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