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Re: Re: Tough Limit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg34283] Re: [mg34251] Re: [mg34235] Tough Limit
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Mon, 13 May 2002 05:54:49 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Strictly speaking you may be right, but it seemed to me pretty clear 
that the questioner meant to take absolute value and from his reply I 
gather that I was right in supposing so.

Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/

On Monday, May 13, 2002, at 02:04  AM, DrBob wrote:

> 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
> Sent: Sunday, May 12, 2002 2:26 AM
> Subject: [mg34283] [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
>

>
> 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
>>
>>
>>
>
>
>
>



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