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

**References**:**Problem with limiits***From:*Roger Bagula <rlbagulatftn@yahoo.com>