Re: abundance of irrationals!)


W. Mueckenheim wrote:
> "Randy Poe" <poespam-trap@xxxxxxxxx> wrote in message
news:<1113499436.071675.26170@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
>
> > Proof that lim(n->oo)1/n = 0:
> >
> > Let eps be any positive value. Define n0 = ceiling(1/eps)
> > so n0 >= 1/eps and 1/n0 <= eps.
> >
> > Then for all n>n0, 1/n < 1/n0 <= eps.
> >
> > Thus |1/n - 0| < eps for all n > n0, and I have established
> > that lim(n->oo)1/n = 0.
> >
> > Note that the statement 1/n < 1/n0 refers to finite values
> > of n. Note that the statement 1/n < 1/n0 is true for all
> > finite values of n with the property that n>n0. Thus,
> > as I said, I have considered only what happens with
> > finite values of n, and by doing so I have proven
> > the limit to be 0.
> >
> > As I said, this limit was evaluated by considering
> > only properties of finite values of n.
>
> Sequences cause no problems, because all n are finite. Series cause
> problems, because N is said to be infinite. Please repeat your proof
> for the infinite Sum 1/2^k = 1, and you will see where you fail.

The definition of this infinite sum is a sequence, and
you said "sequences cause no problems".

But let me show you explicitly.

Let S = sum(k=1,oo) 1/2^k

Theorem: S = 1
Proof:

By definition, S is the limit of the sequence S_1, S_2,
.... where S_n = sum(k=1,n) 1/2^k.

S_n can be evaluated explicitly as a geometric series:

S_n = [(1/2)-(1/2)^(n+1)]/[1 - (1/2)]
= 1 - (1/2)^n

Now consider any eps > 0. For a given value of eps,
choose n0 = ceiling(log(1/eps)/log 2) so that

n0 >= log(1/eps)/log 2
n0 log 2 >= log(1/eps)
2^n0 >= 1/eps

(1/2)^n0 <= eps

Thus for any value of eps we can find n0 such that
(1/2)^n0 <= eps.

Now for every finite n > n0, we have (1/2)^n < (1/2)^n0 <= eps.

Since (1/2)^n = 1 - S_n and (1/2)^n0 = 1 - S_n0, we thus have

0 < 1 - S_n < eps for every finite value of n > n0, or
|1 - S_n| < eps for every finite value of n > n0, and we
can find a suitable n0 for any positive choice of epsilon.

This established that the limit of the sequence S_1, S_2,...
is 1.

By definition, S is the limit of that sequence. Therefore
S = sum(k=1,oo) (1/2)^k = 1.
----------------------------------
Note that the entire argument was based on being able
to find a finite value of n0, and examining what
happens to finite values of n which are greater than
n0. Once I proved the criterion held for every finite
choice of n greater than n0, I was done.

- Randy

.

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