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
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