Re: abundance of irrationals!)

In article <fb701d3c.0504110411.3cfc9932@xxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx (W. Mueckenheim) writes:
> Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx> wrote in message news:<ITSnetNOTcom#virgil-9BE167.17030809042005@xxxxxxxxxxxxxxxxxxxxxxxx>...
>
> > >
> > > 1/2^n is a typical term of the sequence. No term of the sequence "over
> > > all n e N" is equal to the limit.
> > > Why should the above sum over all k e N be equal to the limit?
> >
> > Because for every positive real epsilon, there is a positive integer m
> > such that whenever n >= m then |1 - SUM_{1<=k=<n, 1/2^2| < epsilon.
>
> Whenever n > m means, it must be calculaed for ALL n > m. But a sum
> over all n > m is an infinite sum. This ust be calculaed in order to
> prove that a limit exists. I.e., it must be possible to calculate it
> in advance.

Eh? |1 - sum {k = 1...n} | = 1/2^n, yes? So whenever n >= m that formula
is less then or equal to 1/2^m, yes? And there is no sum over all n > m
involved.
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