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MathGroup Archive 2005

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Re: Re: Summation problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56712] Re: [mg56675] Re: [mg56621] Summation problem
  • From: Devendra Kapadia <dkapadia at wolfram.com>
  • Date: Wed, 4 May 2005 00:34:12 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello Andrzej,

As I mentioned in my earlier email, the wrong behavior in
the infinite sum sent by Jaroslaw Piskorski was caused by
the failure of Sum to detect conditional convergence in this
specific example. This, in turn, was due to a change 
in the output from Limit (which is used by Sum for 
convergence testing), in Mathematica 5.

There was no new method added to Sum in Mathematica 5, and
hence the statement in the documentation referred by you 
seems fine.

Hope this helps. 

Regards,

Devendra Kapadia.
Wolfram Research, Inc.  



On Tue, 3 May 2005, Andrzej Kozlowski wrote:

> *This message was transferred with a trial version of CommuniGate(tm) Pro*
> The documentation for Sum says:
> 
> > New in Version 1; modified in 3.
> 
> 
> Shouldn't it be  "in 3 and 5" ?
> 
> Andrzej Kozlowski
> 
> 
> 
> On 3 May 2005, at 18:26, Devendra Kapadia wrote:
> 
> >
> > On Sat, 30 Apr 2005, jaropis wrote:
> >
> >> Why can't Mathematica sum:
> >> Sum[I^n/n,{n,1,Infinity}]
> >> and gives the (incorrect) answer, that this is divergent while it can 
> >> do:
> >> Sum[I^(n+1)/n,{n,1,Infinity}]
> >> correctly?
> >>
> >> Jaroslaw Piskorski
> >>
> > Hello Jaroslaw Piskorski,
> >
> > Thank you for reporting this problem with an infinite sum.
> >
> > This is caused by a failure to detect the conditional convergence
> > of the sum in your first example, in Mathematica 5.
> >
> > A workaround for the problem is to replace 'I' by the symbolic
> > quantity 'x' and then substitute 'x' with 'I', as shown below.
> >
> > ==========================================================
> > In[1]:= $Version
> >
> > Out[1]= 5.1 for Linux (February 20, 2005)
> >
> > In[2]:= Sum[x^n/n,{n,1,Infinity}]/.{x-> I}
> >
> > Out[2]= -Log[1 - I]
> >
> > In[3]:= N[%]
> >
> > Out[3]= -0.346574 + 0.785398 I
> >
> > In[4]:= NSum[I^n/n,{n,1,Infinity}]
> >
> > Out[4]= -0.346574 + 0.785398 I
> >
> > ============================================================
> >
> > Sorry for the inconvenience caused by this problem.
> >
> > Devendra Kapadia.
> > Wolfram Research, Inc.
> >
> 




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