Writing down harmonic numbers

The nth harmonic number is the sum of the reciprocals of the first n positive integers. Hn = 1 + 1/2 + 1/3 + 1/4 + … + 1/n The product of all the denominators is n!, so you could write Hn as a fraction Hn = p/q where p = n! Hn is an integer and q = n!. While p/q is a way to write Hn as a fraction, it’s not the most efficient because p and n! will have common factors. If we write Hn as a reduced fraction, the denominator will be the least common multiple of the integers 1 through n. That number is asymptotically exp(n). That estimate follows from the prime number theorem. So for large n the denominator will be roughly exp(n), and in base b it would have around n/log(b) digits. The numerator will be exp(n) Hn, and since Hn is asymptotically log(n) + γ, the numerator for large n will be roughly exp(n) (log(n) + γ) and will have around (n + log log(n) ) / log(b) digits. Let’s see how well our asymptotic estimates work for n = 50. The 50th harmonic number is H50 = 13943237577224054960759 /…