Slower summation of a series in Julia than in Python

General Usage Performance
6

timeit.timeit does not work this way. Each call to timeit.timeit returns the time taken to run the command given to timeit.timeit, which in your case with timeit.timeit() is no command. So your main simply returns the difference between two calls to timeit.timeit, each of which gives the time taken to run no command at all. This probably explains why sometimes the difference is negative and you have used abs to ignore that.

To correctly calculate the time taken to run sum_n(harmonic, 500000), you need to do:

import timeit

timeit.timeit("sum_n(harmonic, 500000)", setup="""
def harmonic(k):
   return (-1) ** k / (k + 1)

def sum_n(an, n):
   S = 0.0
   for k in range(n+1):
       S += an(k)
   return S
""", number=1)

Note however that timeit.timeit by default uses many evaluations (a million) of the expression you supply, so the above would take a very long time to return. Use the number argument to timeit.timeit to get an answer in an acceptable time.

Doing that for me gives about 250 ms for the Python code above. So much slower than your original Julia.

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