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MathGroup Archive 2006

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Re: Re: Problem with limiits

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65842] Re: [mg65819] Re: Problem with limiits
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Tue, 18 Apr 2006 06:56:38 -0400 (EDT)
  • References: <e1snmb$860$1@smc.vnet.net> <200604170629.CAA08964@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Roger Bagula wrote:
> Roger Bagula wrote:
> 
>>A well known limit is:
>>Limit[(1 + 1/n)^n, n -> Infinity]=E
>>I tried it and it works... solution seems built in.
>>
>>I tried:
>>Limit[(1 + 1/Prime[n])^Prime[n], n -> Infinity]
>>
>>Again I tried:
>>Limit[(1 + 1/Prime[n])^Prime[n], n -> 2000]
>>
>>Here's how I got an estimate:
>>Table[(1 + 1/Prime[n])^Prime[n], {n, 1, 400}];
>>ListPlot[%]
>>
>>It appears to be approaching  E as well.
>>N[(1 + 1/Prime[2000])^Prime[2000], 100] - E
>>-0.000078156838841603507435562510935842641134579112458192281970712293762387821356624136556497567576200
>>
> 
> It appears that another Limit exists that behaves in the same  way:
> http://mathworld.wolfram.com/Primorial.html
> 
>>The primorial satisfies the unexpected limit
>>lim_(n->infty)(p_n#)^(1/p_n)==e	(3)
>>
>>(Ruiz 1997; Pruitt), where e is the usual base of the natural logarithm. 

"Behaves the same way"? Maybe you mean in the sense of Mathematica 
returning it unevaluated? If you refer to the mathematics, it is a horse 
of a very different color.

For those who do not want to check the MathWorld site, the primordial 
function of n is the product of the first n primes.

We can show that the limit in question is E as below. I do not claim 
that this constitutes a rigorous proof, but it comes close. Details 
would include verification of some computations and validating the use 
of certain approximations.

To show the product is E it suffices to show that the log is 1. This is 
given in Mathematica notation as 1/Prime[n]*Sum[Log[Prime[k]], {k,1,n}]. 
Needless to say, Mathematica does not know a closed form for this.

Now notice that Prime[k] is approximated by k*Log[k]*(1+1/Log[k])

p[k_] := k*Log[k]*(1+1/Log[k])

I used something with ProductLog in an earlier reply to another note in 
this thread, but this approximation is better for most purposes. After 
taking logs we see that the dominating term in the summands is simply 
Log[k]. We now just do some simple manipulations to see that the limit 
in question is 1.

In[35]:= s1 = Sum[Log[k], {k,1,n}]
Out[35]= Log[Gamma[1 + n]]

In[36]:= Limit[s1/p[n], n->Infinity]
Out[36]= 1

So the exponential, which gives the original product in question, is E.


Daniel Lichtblau
Wolfram Research


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