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 > >