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])

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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