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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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