Re: Problem with limiits

*To*: mathgroup at smc.vnet.net*Subject*: [mg65761] Re: [mg65695] Problem with limiits*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 17 Apr 2006 02:27:59 -0400 (EDT)*References*: <200604160544.BAA07913@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 16 Apr 2006, at 14:44, 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 > -0.0000781568388416035074355625109358426411345791124581922819707122937 > 62387821356624136556497567576200 > For any sequence of real numbers a1,a2, a3 .... converging to a real number L, any infinite subsequence of it will converge to the same limit L. This is a very basic fact and is in fact true in any complete metric space (a subsequence of a convergent sequence is a Cauchy sequence, and therefore, in a complete metric space, is itself convergent.) Given the above, you only need the fact that there are infinitely many primes (proved by Euclid and before him by Eudoxus). Andrzej Kozlowski

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