Summation of 1/(n*(n + 1)*(2*n + 1)) , n=1 to infinity converges but symbolically evaluates to nan

[ethankward]

>>> import sympy
>>> sympy.__version__
'1.1.1'
>>> n = sympy.Symbol('n')
>>> s = sympy.Sum(1/(n*(n + 1)*(2*n + 1)), [n, 1, sympy.oo])
>>> s.doit()
nan
>>> sympy.N(s)
0.227411277760219

The actual value is 3 - 4 log(2).

[czgdp1807]

Currently, the code behaves like as follows,

Python 3.6.8 (default, Aug 20 2019, 17:12:48) 
[GCC 8.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from sympy import *
>>> n = symbols('n')
>>> Sum(1/(n*(n + 1)*(2*n + 1)), [n, 1, oo]).doit()
nan
>>> N(Sum(1/(n*(n + 1)*(2*n + 1)), [n, 1, oo]).doit())
nan

[anutosh491]

We do have classes implemented in functions/special/hyper.py to compute the answer though . I believe the answer should be -4*HyperRep_atanh(1) - 2*HyperRep_log1(1) + 3 , but there are some 3-4 errors reported like this which return nan,I am not sure if we have required things to approach those !