Re: abundance of irrationals!)

"*** T. Winter" <***.Winter@xxxxxx> wrote in message news:<IEsD01.6FE@xxxxxx>...
> In article <fb701d3c.0504110500.4882bf50@xxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx (W. Mueckenheim) writes:
> > "*** T. Winter" <***.Winter@xxxxxx> wrote in message news:<IEr5A8.Evx@xxxxxx>...
> ...
> > > > > k in N is meaning. So long as you do not define it it is nothing. So
> > > > > what do you *mean* with the sum over k in N? The standard meaning is:
> > > > > lim{n -> oo} sum {k <= n} 1/2^k.
> > > >
> > > > The standard meaning of Cauchy's convergence criterion is to find out
> > > > something for ALL n > n_0. This implies an infinite sum, but not the
> > > > limit n-->oo the existence of which you want to establish by using
> > > > Cauchy's criterion.
> > >
> > > How do you come at that stupid idea? You have apparently no notion about
> > > the definition of the concept of limits.
> >
> > In order to prove the existence of a limit of a series, you must find
> > S(n) for ALL n in advance. Do you think that all these sums are
> > finite? Do you think that all numbers n form a finite set?
>
> When the terms are 1/2^k, Sn = 1 - 1/2^k. Right? Where is an infinite sum
> involved? And no all numbers n do not form a finite set, but all numbers n
> are finite.

That is why I say that the SUM 1/2^k over all these numbers k e N must
be smaller than 1, because only finite numbers k with 1/2^k > 0 are
involved. Only the finity of the k is important. Not important is, how
many of them may come along. We know: each of them is finite.

Regards, WM
.

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