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Re: Repost --- Another limit problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67858] Re: Repost --- Another limit problem
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Mon, 10 Jul 2006 06:38:44 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <e8iq8s$rvh$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <e8iq8s$rvh$1 at smc.vnet.net>, Virgil Stokes <vs at it.uu.se> 
wrote:

> I am trying to evaluate the limit of the following expression as s goes to 
> infinity,
>  
> \!\(\((1 + \[Sum]\+\(k = 0\)\%\(s - 1\)\((\((s\ \
> \[Rho])\)\^k\/\(k!\))\)/\((\((s\ \[Rho])\)\^s\/\(\(s!\) \((1 - \
> \[Rho])\)\))\))\)\^\(-1\)\)
>  
> where, \[Rho]\ (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.

Writing the expression in InputForm as

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

then, the leading term in the asymptotic expansion can be obtained by 
replacing s-1 by Infinity in the Sum, and using Stirling's formula, 
leading to 

  (r^s E^(s (1 - r)))/(Sqrt[2 Pi s] (1 - r))

Clearly, for 0 < r < 1, this tends to zero as s -> Infinity, since it 
equals

 Exp[s (Log[r] + 1 - r)]/(Sqrt[2 Pi s] (1 - r))

and 

 Log[r] + 1 - r < 0

for 0 < r < 1.

Cheers,
Paul

_______________________________________________________________________
Paul Abbott                                      Phone:  61 8 6488 2734
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