Can SymPy tell me that A = 1/64*pi**2 by simplifying the polylog in below example?
>>> from sympy.abc import n >>> from sympy import summation, oo >>> A = summation(1/((2*n+1)^2-4)^2, (n, 0, oo)) >>> A -polylog(2, exp_polar(I*pi))/16 + pi**2/96 >>> A.simplify() -polylog(2, exp_polar(I*pi))/16 + pi**2/96 because it seems doable: 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 This came up in https://ask.sagemath.org/question/35839/sage-incorrectly-evaluates-series/ Sébastien -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/1f93f41c-ebf8-4402-990d-cc4f9ca22aa5%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
