I asked this question myself a few months ago, and the easiest 2
solutions seem to be utilizing sympy or maxima.
Via sympy it is:
import sympy
sympy.var('x')
print sympy.sum(2**(-x), (x, 1, oo))
I'm taking this from a question I posed on the sympy message list:
http://groups.google.com/group/sympy/browse_frm/thread/5348ded3ebe8a25e?tvc=1
It should return a result of 1, but in sage 3.1.1 it returns 1-2*2**(1-
Infinity). While technically correct, this should clearly return 1
when simplified so I guess there is some complication when
transferring between modules. Ondrej will assuredly give more useful
information if he sees this.
More specifically , your example using m=2 is:
sympy.sum(1/((x+2)**3)),(x,1,oo))
Unfortunately this returns
Sum((2 + x)**(-3), (x, 1, Infinity))
n() on this function does not work, maybe a sympy equivalent would?
It may work better with a %python header, though I haven't tested this
yet (if you use the notebook).
I don't remember the maxima parsing offhand. I believe I got some
information about it from delving into the sage/washington undergrad
mail list. I'll try to look into it tomorrow. Making a natural
implementation for infinite series seems quite valuable and hopefully
a short-term goal considered for Sage. This is a specific dismay I've
come across when trying to broadcast Sage to an otherwise quite pro-
opensource professor.
Would a function called nsum or such that called sympy/maxima be
feasible? This would lead to a supplementation rather than a
replacement for the python sum().
Thomas
On Aug 29, 5:54 pm, Raouf <[EMAIL PROTECTED]> wrote:
> Hi,
> I am a newbie in sage and i want to compute an infinite sum with
> parameter m, like sum(1/(k+m)^3) k=1 to infinity.
> Can you please help me?
> thanks
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