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Re: Another limit problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67766] Re: [mg67711] Another limit problem
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 6 Jul 2006 06:54:52 -0400 (EDT)
  • References: <200607050818.EAA26450@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 5 Jul 2006, at 17:18, Virgil Stokes wrote:

> I am trying to evaluate the limit of the following expression as s  
> goes
> to infinity,
>
> \!\(\((1 + â??\+\(k = 0\)\%\(s - 1\)\((\((s\ Ï?)\)\^k\/\(k!\))\)/\((\ 
> ((s\ \
> Ï?)\)\^s\/\(\(s!\) \((1 - Ï?)\)\))\))\)\^\(-1\)\)
>
> where, Ï? (Real) < 1, s (Integer) > 0. I am quite sure that the  
> limit is
> 0; but, I am unable to get this result using Mathematica 5.2.
> Any suggestions would be appreciated.
>
> --V. Stokes
>


I don't think Mathematica alone can prove this but with the help of  
Mathematica I seem to be able to prove it for any numerical rho, and  
probably could do so in general if I could devote a little more time  
to it. But as I can't I have decided to post my incomplete argument  
below, hoping that someone will either to complete the proof or find  
my mistake (or perhaps a bug in Mathematica ;-)).

  All we need to show that the reciprocal of your expression goes to  
Infinity as s->Infinity. Mathematica gives us the following  
expression in terms of incomplete Gamma:


FunctionExpand[FullSimplify[
    Sum[(s*r)^k/((k!*(s*r)^s)/
       (s!*(1 - r))),
     {k, 0, s - 1}]]]


((-E^(r*s))*(r - 1)*s*
    Gamma[s, r*s])/(r*s)^s

where I used r instead of your rho (and I ignored +1 as it does not  
affect the answer). Now let's us re-write it in the form:

s!*((E^(r*s)*(1 - r))/
    (r*s)^s)*(Gamma[s, r*s]/
    (s - 1)!)

 From the formula for the value of Gamma[n,x] where n is an integer  
given here:

http://mathworld.wolfram.com/IncompleteGammaFunction.html

we see that  Gamma[s, r*s]/(s - 1)!) -> 1 as s->Infinity. Therefore  
our limit is the same as the limit

Limit[s!*(E^(r*s)*(1 - r) /
    (r*s)^s),s->Infinity,Assumptions ->{0<r<1}]

Unfortunately Mathematica 5.1 is still unable to resolve this, but if  
we substitute a numerical value for r it seems to be able to do so,  
and the answers it gives seem to be as follows:

0<r<1


r = 3/4; Limit[
    s!*(E^(r*s)*((1 - r)/
       (r*s)^s)),
    s -> Infinity,
    Assumptions -> {0 < r < 1}]

Infinity


1<r

r = 2; Limit[s!*(E^(r*s)*
      ((1 - r)/(r*s)^s)),
    s -> Infinity]


-Infinity

It suggests that your statement is actually valid for any r>0. I  
think this should not be too hard to prove by hand but I have  
absolutely no more time to think about it as I am leaving for Europe  
tomorrow morning and must start packing now! If the problem is still  
open in about one week's time I will try to return to it.

Andrzej Kozlowski

Tokyo, Japan


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