Re: abundance of irrationals!)

In article <fb701d3c.0504110456.5298f0df@xxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx (W. Mueckenheim) writes:
> "*** T. Winter" <***.Winter@xxxxxx> wrote in message news:<IEr61u.GHJ@xxxxxx>...
> > In article <IEr5A8.Evx@xxxxxx> "*** T. Winter" <***.Winter@xxxxxx> writes:
> > > In article <fb701d3c.0504091328.686ed77a@xxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx (W. Mueckenheim) writes:
> > ...
> > > > You oppose to infinite sums but you need a proof for ALL n > n_0 which
> > > > ultimately leads to an infinite sum, before you know whether the
> > > > series is converging.
> > >
> > > How wrong you are. Infinite sums do *not* exist (there is no theorem
> > > or axiom that gives meaning to them). In the case:
> > > sum{k=1...n} 1/2^k = 1 - 1/2^k
> > > so by the *definition* of limit the series converges and its limit is 1.
> > > Also you do not understand Cauchy's convergence criterion. It does not
> > > talk about sums, but only about differences between pairs of elements.
> >
> > Sorry, this is wrong (of course). It talks about differences of finite
> > sums.
>
> It talks about "ALL n" > n_0. Do you think "ALL n" supply a finite sum?

Yes, because all n are finite. So for which n do you get an infinite sum?
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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