Re: Re: Problem with limiits

*To*: mathgroup at smc.vnet.net*Subject*: [mg65858] Re: [mg65845] Re: Problem with limiits*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 19 Apr 2006 04:54:07 -0400 (EDT)*References*: <200604160544.BAA07913@smc.vnet.net> <e1vdsq$967$1@smc.vnet.net> <200604181056.GAA14316@smc.vnet.net> <D4C101CB-009C-4E4D-B69F-0550E55C075D@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

I have to add one more thing. In my original reply I mentioned complete metric spaces and Cauchy sequences but all of that is quite unnecessary. In fact it is obvious form the definition of convergence that a subsequence of a convergent sequence converges to the same limit as the original sequence. The prof is trivial but if anyone still doubts it then can consult the well known textbook of Rudin, "Principles of Mathematical Analyis", page 51. The author states this fact but considers it too trivial to bother with a proof so he leaves the details to the reader. (I don't know why I thought of the argument using Cauchy sequences - true but completely unnecessary - I guess I must have been thinking of some other related, less trivial result). That Table[(1 + 1/Prime[n])^Prime[n], {n, 1, Inifinity}] is a subsequence of Table[(1 + 1/m)^m, {m, 1, Infinity}] is, if anything, even more trivial. Andrzej Kozlowski On 18 Apr 2006, at 21:53, Andrzej Kozlowski wrote: > > > On 18 Apr 2006, at 19:56, Roger Bagula wrote: > >> Andrzej Kozlowski wrote: >> >>> >>> 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 >>> >> What you are saying is that if I have a composite number c[m] >> them the Limit: >> >> Limit[(1+1/c[m])^c[m],m->Infinity]=E >> >> as well. >> What About: >> Limit[(1-1/m)^m,m->Infinity] =1/E >> >> Your reasoning gives: >> Limit[(1-1/c[n])^c[n],n->Infinity]*Limit[(1+1/Prime[n])^Prime[n],N- >> >Infinity]=1 >> and >> Limit[(1-1/m)^m,m->Infinity]*Limit[(1+1/m)^m,m->Infinity]=1 >> This does give: >> Limit[(1 - 1/m^2)^m, m -> Infinity]=1 >> >> Thanks for your help. >> > > No, I am saying that you have to have a *sub-sequence*. > > (1+1/Prime[n])^Prime[n] for n from 1 to Infinity is a sub-sequence > of (1 + 1/m)^m from 1 to Infinity. If you can't see this general > fact just look at > > > Table[(1 + 1/Prime[n])^Prime[n], {n, 1, 3}] > > > {9/4, 64/27, 7776/3125} > > and > > > > > > {2, 9/4, 64/27, 625/256, 7776/3125} > > and it should become clear. > > On the other hand: > > (1-1/m)^m for m from 1 to Infinity is not a sub-sequence of (1 + 1/ > n)^n from 1 to Infinity. > > Just look at the first few terms: > > > Table[(1 - 1/m)^m, {m, 1, 5}] > > > {0, 1/4, 8/27, 81/256, 1024/3125} > > > Do I need to say more? > > Andrzej Kozlowski > >

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

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

**Re: Re: Problem with limiits**

**Re: Re: unable to FullSimplify**

**Re: Re: Problem with limiits**

**Re: Problem with limiits**