Ask Your Question

# Sage incorrectly evaluates series

It incorrectly evaluates $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}=\frac{\pi^2}{64}-\frac{1}{12}$, but correct answer is $\displaystyle\frac{\pi^2}{64}$

edit retag close merge delete

## Comments

Thanks for reporting !

( 2016-12-01 07:57:47 -0500 )edit

## 1 answer

Sort by » oldest newest most voted

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

more

## Comments

Thanks! We should also file an upstream bug report at Maxima - can you do that as well?

( 2016-12-01 06:26:30 -0500 )edit

I do not not have a account on the sourgeforge of maxima. Also I do not have access to version latest version 5.38. If possible, I will let somebody else do the upstream bug report.

( 2016-12-01 06:55:09 -0500 )edit

Thanks for the ticket !

( 2016-12-01 07:59:13 -0500 )edit

This may actually be https://sourceforge.net/p/maxima/bugs... which is apparently fixed in upstream.

( 2016-12-01 08:04:19 -0500 )edit

I've got ValueError: Mathematica cannot make sense of input sum(1/((2*x+1)^2-4)^2,x,0,Infinity, algorithm='mathematica')

( 2016-12-01 09:11:02 -0500 )edit

## Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

## Stats

Asked: 2016-12-01 02:50:57 -0500

Seen: 97 times

Last updated: Dec 01 '16