After contacting the Maxima team it turns out that Maxima indeed can
evaluate the inifinite sum (at least a numerical value but not a closed
form which probably does not exist)
Sage:
k=var('k'); sum(log(1-2^-k),k,1,20) results in an error message ending with
"Sum is divergent".
Maxima:
load
A bit of exploration in the respective interpreters show that neither
Maxima, Sage, Giac, Fricas, Sympy nor Mathematica can give an expression
for this sum. Wolfram Alpha gives the numerical answer the OP reported, but
I have been unable to get any explanation about the process used ; graphics
Investigating the error shows that sympy didn't do anything with this sum:
671 try:
--> 672 return result._sage_()
673 except AttributeError:
674 raise AttributeError("Unable to convert SymPy result
(={}) into"
ipdb> p result
Sum(log(1 - 1/2**
Thanks for investigating. I tried to use sympy this way:
sum(log(1-2^-k),k,1,oo,algorithm="sympy")
but this failed for other reasons (NotImplementedError: conversion to
SageMath is not implemented)
wdjo...@gmail.com schrieb am Dienstag, 10. Dezember 2024 um 14:58:49 UTC+1:
> On Tue, Dec
On Tue, Dec 10, 2024 at 8:43 AM 'OHappyDay' via sage-support <
sage-support@googlegroups.com> wrote:
> I tried to evaluate the infinite product:
>
> prod((2^n-1)/2^n) (n=1,oo)
>
> by converting the product to a sum via logarithm:
>
> sum(log(1-2^-k),k,1,oo)
>
> The sum (and thus the product) shoul
I tried to evaluate the infinite product:
prod((2^n-1)/2^n) (n=1,oo)
by converting the product to a sum via logarithm:
sum(log(1-2^-k),k,1,oo)
The sum (and thus the product) should, according to WolframAlpha, converge
with a final value of about
-1.24206
This video (https://www.youtube.com/w
