# Revision history [back]

Indeed:

sage: n = var('n')
sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity)
1/64*pi^2 - 1/12


Mathematica does it correctly:

sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity, algorithm='mathematica')
1/64*pi^2


Giac gives this:

sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity, algorithm='giac')
1/32*Psi(-1/2, 1) - 1/8


And SymPy seems to do it correctly:

sage: from sympy.abc import n
sage: from sympy import summation, oo
sage: A = summation(1/((2*n+1)^2-4)^2, (n, 0, oo))
sage: A._sage_()
1/64*pi^2


I created ticket #22004 so that one can do:

sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity, algorithm='sympy')


Indeed:

sage: n = var('n')
sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity)
1/64*pi^2 - 1/12


See ticket #22005. Mathematica does it correctly:

sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity, algorithm='mathematica')
1/64*pi^2


Giac gives this:

sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity, algorithm='giac')
1/32*Psi(-1/2, 1) - 1/8


And SymPy seems to do it correctly:

sage: from sympy.abc import n
sage: from sympy import summation, oo
sage: A = summation(1/((2*n+1)^2-4)^2, (n, 0, oo))
sage: A._sage_()
1/64*pi^2


I created ticket #22004 so that one can do:

sage: sum(1/((2*n+1)^2-4)^2, n, 0, Infinity, algorithm='sympy')
1/64*pi^2