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Re: Limit[(p*(-1 + p^n))/(-1 + p), n->Infinity]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28928] Re: [mg28918] Limit[(p*(-1 + p^n))/(-1 + p), n->Infinity]
  • From: andrzej <andrzej at bekkoame.ne.jp>
  • Date: Sat, 19 May 2001 22:27:48 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

There is one trick, which often works in such cases and which I have 
described several times in this group in the past. The idea is to 
replace the limit of a sequence a[n] by the infinite sum 
Sum[a[n+1]-a[n],{n,1,Infinity}]+a[1]. Mathematica can quite frequently 
cope with such expressions symbolically. This is one of these cases:

In[2]:=
Simplify[Sum[(p*(p^(n + 1) - 1))/(p - 1) -
      (p*(p^n - 1))/(p - 1), {n, 1, Infinity}] + p]

Out[2]=
   p
-----
1 - p

Note that Mathematica here makes a silent assumption that -1<p<1. It 
sees to me somewhat questionable if such silent assumptions are a good 
thing (I would prefer answers in the form If[....]) but this happens 
quite often in Mathematica and seems to me to be a legacy from early 
versions (before Assumptions and Domains).

Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
http://sigma.tuins.ac.jp/~andrzej/


On Friday, May 18, 2001, at 02:13  PM, Adalbert Hanssen wrote:

> Hi, MathGroup,
>
> of course, this one is easily done by hand:
>
> Limit[(p*(-1 + p^n))/(-1 + p), n->Infinity]
>
> but how does one do it with Mathematica?
> Of course, 0<=p<=1 (real) and n runs through integers.
>
>
> kind regards
>
> Dipl.-Math. Adalbert Hanszen
>
>


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