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

@recursive's solution is correct for a floating point approximation. If you prefer, you can get the exact answer in Python 3.0 using the fractions module:

>>> from fractions import Fraction
>>> def calc_harmonic(n):
...   return sum(Fraction(1, d) for d in range(1, n + 1))
...
>>> calc_harmonic(20) # sum of the first 20 terms
Fraction(55835135, 15519504)

Note that the number of digits grows quickly so this will require a lot of memory for large n. You could also use a generator to look at the series of partial sums if you wanted to get really fancy.

Answer 2

The harmonic series diverges, i.e. its sum is infinity..

edit: Unless you want partial sums, but you weren't really clear about that.

Answer 3

How about this:

partialsum = 0
for i in xrange(1,1000000):
    partialsum += 1.0 / i
print partialsum

where 1000000 is the upper bound.

Answer 4

Homework?

It's a divergent series, so it's impossible to sum it for all terms.

I don't know Python, but I know how to write it in Java.

public class Harmonic
{
    private static final int DEFAULT_NUM_TERMS = 10;

    public static void main(String[] args)
    {
        int numTerms = ((args.length > 0) ? Integer.parseInt(args[0]) : DEFAULT_NUM_TERMS);

        System.out.println("sum of " + numTerms + " terms=" + sum(numTerms));
     }

     public static double sum(int numTerms)
     {
         double sum = 0.0;

         if (numTerms > 0)
         {
             for (int k = 1; k <= numTerms; ++k)
             {
                 sum += 1.0/k;
             }
         }

         return sum;
     }
 }

Answer 5

By using the numpy module, you can also alternatively use:

import numpy as np
def HN(n):
    return sum(1/arange(1,n+1))

Answer 6

I add another solution, this time using recursion, to find the n-th Harmonic number.

General implementation details

Function Prototype: harmonic_recursive(n)

Function Parameters: n - the n-th Harmonic number

Base case: If n equals 1 return 1.

Recur step: If not the base case, call harmonic_recursive for the n-1 term and add that result with 1/n. This way we add each time the i-th term of the Harmonic series with the sum of all the previous terms until that point.

Pseudocode

(this solution can be implemented easily in other languages too.)

harmonic_recursive(n):
    if n == 1:
        return 1
    else:
        return 1/n + harmonic_recursive(n-1)

Python code

def harmonic_recursive(n):
    if n == 1:
        return 1
    else:
        return 1.0/n + harmonic_recursive(n-1)