SymPy doit returns NaN instead of zero

[Hadas Biran]

I'm trying to evaluate a long expression with lots of summations and n choose k's. For some reason, the following expression (which is a part of the long expression) returns NaN.

from sympy import *

w_1_2_3 = symbols('w_1_2_3')

w_1_3 = symbols('w_1_3')

w_2_3 = symbols('w_2_3')

w_1_2 = symbols('w_1_2')

exprr = 7*Sum(binomial(5, w_1_2_3)*Sum(binomial(0, -w_1_2_3 + w_1_3)*binomial(2, w_1_2_3 - w_1_3 - 2)*binomial(13, 11 - w_1_2_3), (w_1_3, w_1_2_3, 5)), (w_1_2_3, 0, 5))/868017280

exprr.doit() -> NaN

However, if I evaluate each of the terms in the first sum separately, I get 0 for each of them (as should be):

exprr2 = binomial(5, w_1_2_3)*Sum(binomial(0, -w_1_2_3 + w_1_3)*binomial(2, w_1_2_3 - w_1_3 - 2)*binomial(13, 11 - w_1_2_3), (w_1_3, w_1_2_3, 5))

for t in range(0, 6):

      res = exprr2.subs(w_1_2_3, t).doit()

      print(t, res) 

-> 0 0 

     1 0 

     2 0 

     3 0 

     4 0 

     5 0

What is going on here? Why do I get NaN for the overall term?

I also posted a question in StackOveflow-https://stackoverflow.com/questions/73222278/sympy-doit-returns-nan-instead-of-zero

[Aaron Meurer]

I'd say this is a bug. The problem it it's evaluating the inner sum to
an expression that has terms like (w_1_2_3 - 1) in both the numerator
and denominator without cancelling them. You can work around it by
evaluating and simplifying the inner summand separately:

>>> 7*Sum((binomial(5, w_1_2_3)*Sum(binomial(0, -w_1_2_3 + w_1_3)*binomial(2, w_1_2_3 - w_1_3 - 2)*binomial(13, 11 - w_1_2_3), (w_1_3, w_1_2_3, 5)).doit()).simplify(), (w_1_2_3, 0, 5)).doit()/868017280
0

Aaron Meurer

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