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MathGroup Archive 2006

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Re: Problem with limiits

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65838] Re: [mg65695] Problem with limiits
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Tue, 18 Apr 2006 06:56:34 -0400 (EDT)
  • References: <200604160544.BAA07913@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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.

Not sure what this means. This limit is computed via a series expansion.


> I tried:
> Limit[(1 + 1/Prime[n])^Prime[n], n -> Infinity]

Limit in Mathematica works with functions defined on a continuum in a 
one-sided open neighborhood of the point in question. In this case that 
would be something of the form (k,Infinity) where k is some finite 
value. Prime[n] is only defined on a discrete set. On functions of that 
sort Limit will only "work" by accident.


 > 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

Again the Series expansion at infinity of (1 + 1/k)^k is revealing. The 
first order error is -E/(2*k). Now use an approximation to k=Prime[n] as 
-n*ProductLog[-1,-1/n] to get an estimate of error as 
E/(2*n*ProductLog[-1,-1/n]).


Daniel Lichtblau
Wolfram Research


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