RE: Re: Tough Limit
- To: mathgroup at smc.vnet.net
- Subject: [mg34274] RE: [mg34251] Re: [mg34235] Tough Limit
- From: "DrBob" <majort at cox-internet.com>
- Date: Mon, 13 May 2002 05:54:19 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
Taking the absolute value changes the problem, not the answer.
The following:
FullSimplify[Binomial[-1/2, n]
/Binomial[2n,n],Element[n,Integers]]//InputForm
evaluates to
(-1/4)^n
confirming the identity you mentioned below.
So... I let
a=Binomial[-1/2,n]
b=((-1/4)^n*Gamma[1 + 2*n])/Gamma[1 + n]^2
c=b/.Gamma[1+n_]->n!
d=c/.k_!->(k^(1/2 + k)*Sqrt[2*Pi])/E^k
e=d Sqrt[n]*Sqrt[Pi]
Skipping most of the outputs, the last one evaluates to:
(-1)^n
This has no limit as n tends to infinity (unless you change the problem
by taking the absolute value).
An earlier answer from Vladimir Bondarenko that drew plots to show the
series is all over the interval {-1, 1} was misleading because the plots
aren't restricted to integer n. (MY earlier answer was even MORE bogus.
Mea culpa!!)
Bobby Treat
-----Original Message-----
From: Andrzej Kozlowski [mailto:andrzej at platon.c.u-tokyo.ac.jp]
To: mathgroup at smc.vnet.net
Subject: [mg34274] [mg34251] Re: [mg34235] Tough Limit
First of all, you need to take the absolute value here, since your
expression is negative for odd n, e.g.
In[3]:=
((Binomial [-1/2 ,n ] * Sqrt[ n* Pi ])/.n->99)//N
Out[3]=
-0.998738
I don't think Mathematica can do solve this problem without a lot of
human human help. Here is one way that makes just a slight use of
Mathematica (for the sake of decency).
First we use the identity Binomial[-1/2, n] == (-1/4)^n Binomial[2n, n],
which unfortunately Mathematica does not know but you can find it
somewhere in Graham, Knuth, Patashink, "Concrete Mathemaitcs" together
with about a million other formulas of this type. Hence Abs[Binomial
[-1/2, n]] == (2n)!/(4^n (n!)^2). Now we use the assymptotic Sterling
formula (also proved in the same reference) and let Mathematica do the
cancellation:
In[4]:=
Sqrt[n*Pi]*((2*n)!/(4^n*n!^2)) /. k_! -> Sqrt[2*Pi*k]*(k/E)^k
Out[4]=
1
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
On Saturday, May 11, 2002, at 05:04 PM, RJMilazzo wrote:
> Can anyone suggest how I can use Mathematica to get the following
limit:
>
> limit ( Binomial [-1/2 ,n ] * Sqrt[ n* Pi ] ) as n-> Infinity
>
> I have tried both Calculus`Limit` and the standard Limit functions. I
> can
> verify with NLimit that this limit equals approximately 1. I don't
> think that
> this is rigorous enough for a proof.
>
> Thanks
> James
> rjmilazzo at aol.com
>
>
>