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Re: Another limit problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67769] Re: [mg67711] Another limit problem
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 6 Jul 2006 06:55:08 -0400 (EDT)
  • References: <200607050818.EAA26450@smc.vnet.net> <443A4967-EC98-468B-9015-86ACE77D7987@mimuw.edu.pl> <7F1F9E54-C0B3-4010-95F8-9C41E024D347@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

On 6 Jul 2006, at 11:05, Andrzej Kozlowski wrote:

> Immediately after posting this I realized I had done something  
> silly (which however does not affect the main point). There is  
> completely no need to use Mathematica's FullSimplify and  
> FunctionExpand and the reference to the MathWorld site to get to  
> the expression
>
> Limit[s!*(E^(r*s)*(1 - r) /
>    (r*s)^s),s->Infinity,Assumptions ->{0<r<1}]
>
> In fact this can be done by hand just by looking at the original  
> problem as posted by Virgil. The rest of what I wrote, I think, is  
> O.K.
>
> Andrzej Kozlowski
>
>
> On 6 Jul 2006, at 10:51, Andrzej Kozlowski wrote:
>
>>
>> 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
>

It seems to me that I can now complete the proof. It remains to show  
that

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

By Stirling's approximation we can replace s! by s^(s+1/2)*Exp[-s].  
This gives the following Limit:


Limit[s^(s+1/2)*Exp[-s]*(E^(r*s)*(1 - r) /
    (r*s)^s),s->Infinity,Assumptions ->{0<r<1}]

Canceling and removing the irrelevant factor 1-r we need to determine

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

Mathematica still can't do it, but if we can prove that E^(r - 1)/r) 
 >=1 for 0<r<1 then the result will follow. But this is obvious since  
1 is actually the minimum value of this function on the interval {0,1}.

So it seems that we have a complete strongly suggest  checking it  
carefully before trusting it.

Andrzej Kozlowski


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