Re: abundance of irrationals!)

In article <fb701d3c.0504130312.3c3b96e5@xxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx (W. Mueckenheim) wrote:

> richard@xxxxxxxxxxxxxxx (Richard Tobin) wrote in message
> news:<d3e1jn$2udj$1@xxxxxxxxxxxxxxxxxxxxxxx>...
> > In article <fb701d3c.0504110456.5298f0df@xxxxxxxxxxxxxxxxxx>,
> > W. Mueckenheim <mueckenh@xxxxxxxxxxxxxxxxx> wrote:
> >
> > >It talks about "ALL n" > n_0. Do you think "ALL n" supply a finite sum?
> >
> > Yes. For all n, the sum is finite, because all for all n, n is finite.
>
> That's why I said that
> SUM 1/2^k < 1
> k e N
>
> I is not important, how many of the terms 1/2^k may come along. W know
> for sure that each of them is > 0, so that 1 - 1/2^k remains strictly
> < 1.

What everyone else understands by your symbol is
LIMIT_{n -> oo} ( SUM_{ k=1..n} 1/2^k )
since it has no legitimate mathematical meaning otherwise.
.