Python program to calculate harmonic series

Question

Does anyone know how to write a program in Python that will calculate the addition of the harmonic series. i.e. 1 + 1/2 +1/3 +1/4...

Answer 1

Just a footnote on the other answers that used floating point; starting with the largest divisor and iterating downward (toward the reciprocals with largest value) will put off accumulated round-off error as much as possible.

Answer 2

A fast, accurate, smooth, complex-valued version of the H function can be calculated using the digamma function as explained here. The Euler-Mascheroni (gamma) constant and the digamma function are available in the numpy and scipy libraries, respectively.

from numpy import euler_gamma
from scipy.special import digamma

def digamma_H(s):
    """ If s is complex the result becomes complex. """
    return digamma(s + 1) + euler_gamma

from fractions import Fraction

def Kiv_H(n):
    return sum(Fraction(1, d) for d in xrange(1, n + 1))

def J_F_Sebastian_H(n):
    return euler_gamma + log(n) + 0.5/n - 1./(12*n**2) + 1./(120*n**4)

Here's a comparison of the three methods for speed and precision (with Kiv_H for reference):

Kiv_H(x) J_F_Sebastian_H(x) digamma_H(x) x seconds bits seconds bits seconds bits 1 5.06e-05 exact 2.47e-06 8.8 1.16e-05 exact 10 4.45e-04 exact 3.25e-06 29.5 1.17e-05 52.6 100 7.64e-03 exact 3.65e-06 50.4 1.17e-05 exact 1000 7.62e-01 exact 5.92e-06 52.9 1.19e-05 exact

Answer 3

Using the simple for loop

def harmonicNumber(n):
x=0
for i in range (0,n):
    x=x+ 1/(i+1)
return x

Answer 4

@Kiv's answer is correct but it is slow for large n if you don't need an infinite precision. It is better to use an asymptotic formula in this case:

#!/usr/bin/env python
from math import log

def H(n):
    """Returns an approximate value of n-th harmonic number.

       http://en.wikipedia.org/wiki/Harmonic_number
    """
    # Euler-Mascheroni constant
    gamma = 0.57721566490153286060651209008240243104215933593992
    return gamma + log(n) + 0.5/n - 1./(12*n**2) + 1./(120*n**4)

@Kiv's answer for Python 2.6:

from fractions import Fraction

harmonic_number = lambda n: sum(Fraction(1, d) for d in xrange(1, n+1))

Example:

>>> N = 100
>>> h_exact = harmonic_number(N)
>>> h = H(N)
>>> rel_err = (abs(h - h_exact) / h_exact)
>>> print n, "%r" % h, "%.2g" % rel_err
100 5.1873775176396242 6.8e-16

At N = 100 relative error is less then 1e-15.

Answer 5

This ought to do the trick.

def calc_harmonic(n):
    return sum(1.0/d for d in range(2,n+1))