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 | 1 | initial version |
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')
| | 2 | No.2 Revision |
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
