Re: Tough Limit
- To: mathgroup at smc.vnet.net
- Subject: [mg34251] Re: [mg34235] Tough Limit
- From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
- Date: Sun, 12 May 2002 03:25:48 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
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 > > >