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MathGroup Archive 2004

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Re: infinite sum problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52992] Re: [mg52990] infinite sum problem
  • From: Andrzej Kozlowski <andrzej at akikoz.net>
  • Date: Sun, 19 Dec 2004 06:14:35 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

On 18 Dec 2004, at 18:00, marc wrote:

> Hi,
>
> the following sum converges only for x<1,
> but Mathematica gives a result for general x
> when the sum is evaluated symbolically.
>
>
> In[1]:=
> Sum[(-1)^k*x^(2k),{k,0,Infinity}]
>
> Out[1]=
> \!\(1\/\(1 + x\^2\)\)
>
> In[2]:=
> Sum[(-1)^k*x^(2k),{k,0,Infinity}]/.x\[Rule]2
>
> Out[2]=
> \!\(1\/5\)
>
> that's wrong!
>
> In[3]:=
> Sum[(-1)^k*2^(2k),{k,0,Infinity}]
>
> correctly gives error: Sum does not converge.
>
>
> Am I overlooking something?
>
> thanx,
> marc

No, you are not overlooking anything nor is there anything wrong here, 
except your substitution. Sum does not as yet accept any assumptions 
and it uses the principle that used to be almost universal in 
Mathematica before the Assumptions mechanism was introduced: which 
means it returns a convergent solution if one exists and leaves to you 
the job of finding the region of convergence. However, in version 5.1 
Limit works with Assumptions so you can instaed do:


Limit[Sum[(-1)^k*x^(2*k), {k, 0, m}], m -> Infinity,
   Assumptions -> {Abs[x] < 1}]

1/(x^2 + 1)


Limit[Sum[(-1)^k*x^(2*k), {k, 0, m}], m -> Infinity,
   Assumptions -> {Abs[x] > 1}]


ComplexInfinity



Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/~andrzej/
http://www.mimuw.edu.pl/~akoz/


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