So at the end of last time we'd just been looking at series in
the extended real line, and seen that they behave very well.
We'd also observed that
that looks like there was some sort of a problem
with the algebra of limits, which we will revisit now.
So we're going to have a look at the extent to which
the algebra of limits does work in the extended real line,
especially the non-negative extended real numbers.
Now, as with most of the stuff here,
I won't go through many of the proofs in great detail yet.
I'm going to start to do proofs in a bit more detail when we get to
collections of sets, when
the material gets a bit more abstract, and a bit harder
and I'll do you some proper proofs then, but here:
algebra of limits,
we already know that works in the real line, for
if you've got two convergent sequences of real numbers,
then you can add them term by term
and get another convergent sequence of real numbers, and the limit is
what you expect, and
the same for multiplying them term by term.
So now you try the same thing in the extended real line.
And, well, there will be some issues if you were working
with negative extended real numbers,
because you're not allowed to add plus infinity to minus infinity.
But we're not doing that. We're working, just for the moment,
at least, we're working only
in the non-negative extended real numbers, so there's
no problems with addition.
And as a result, there's actually no problem at all
with the algebra of limits for sums
either, it's only products, here, where the problems arise.
So if you want to use
the usual algebra of limits to deduce the new algebra of limits,
what happens is you've just got a few extra cases to consider.
And I've almost given you enough information already
with that Sandwich Theorem I gave you last time.
Essentially, if you've got two sequences of
extended, non-negative extended real numbers
and they both converge to a real number,
then what actually happens is, after a while,
both the sequences will consist
only of real numbers, and you'll just get
back to the usual algebra of limits, for real sequences.
But if either of these sequences tends to plus infinity,
then x_n + y_n will be at least as big as x_n,
and at least as big as y_n,
so if one of those two sequences goes off to infinity,
then the bigger one has to as well,
so you just get it directly from the sandwich theorem.
So it's sort-of two cases.
1) Either of the sequences
tends to infinity, then you get it from
the Sandwich Theorem. Case 2)
both of the sequences converge to
a finite, to some non-negative real number.
In that case, you say, well, okay,
there might be a few infinities in there, but it's got to leave
infinity eventually, and stay away from infinity.
And then you just get ...
you look at what's left of the sequences,
and it's just two convergent sequences of real numbers.
And you already know about those,
you're allowed to quote the result for real numbers.
So it's just again, a little bit of
a boost up, and it works, and it's fine.
So that's no problem.
You can check the details
as an exercise.
So the problems come for products
and we already looked at 1 over n,
(oh, let's have a different colour), we already
looked at (one over n) times n equals one,
which tends to one as n tends to infinity.
But 1/n tends to 0 and n tends to infinity,
and 0 times infinity equals 0, okay?
So that's one problem. Also
infinity times one ... times (one over n) equals infinity,
and that tends to infinity, is not equal to infinity times 0.
So why's that a problem?
Well, if you think of infinity as
your alpha, or as a constant sequence,
whichever, and your one over n as tending to 0,
then your algebra of limits has gone wrong again,
so even constant multiples of sequences go wrong
if your constant multiple is infinity.
And even if you aren't multiplying by infinity
directly, as in (one over n) times n,
you end up with a problem in the limit with 0 times infinity.
So the algebra of limits for products doesn't quite work as it stands.
The ... as long as your alpha isn't infinity,
you can do constant multiples and you're OK. That's the only,
the only problem with constant multiples
is if alpha is infinity: alpha equals 0 is a special case,
but fortunately 0 always wins, so you're OK.
So you can check that all the other
alphas except for infinite alphas are okay.
But there is a problem, and the question is how do you fix it?
And the answer is that you have to
restrict attention to which sequences you want to do
the algebra of limits for, when you're dealing with products.
So we restrict attention to nondecreasing sequences.
So what are we going to do? We're going to have,
we're going to have x_1, so 0 less than or equal to x_1
less than or equal to x_2 less than or equal to x_3, and so on.
And that ... any non-decreasing sequence has to converge to
something. And you have the same for y.
Then the algebra of limits now does work,
and you get that x_n y_n tends to xy.
Of course you also have that it's non-decreasing, so you have
0 less than or ... let's get this right
0 less than or equal to x_1 y_1 less than or equal to x_2 y_2,
and it does tend to xy.
So the algebra of limits is okay,
as long as you're careful.
You can't do any old monotone sequences
because we've ... 1 over n was
monotone, but decreasing, and gave us
trouble, even with the constant sequence infinity,
which is also monotone decreasing, or nonincreasing.
So if you're working with non-negative extended real numbers,
you should work with
nondecreasing sequences, monotone increasing sequences.
If you're going to work with negative extended real numbers instead,
or non-positive, extended real numbers,
then you would want to work with
decreasing sequences (if you want to deal with products) instead.
And that's how that would work.
Well, it's an exercise to check it,
but I'll give you some hints.
Here are some cases.
If neither x_n nor y_n tends to infinity,
then it's just the algebra of limits in R.
Right, so if neither of them tend to infinity,
then it follows from the algebra of limits in R,
because they're all real numbers.
Usually what you would say is
if they tend to real number, then after a while they are real numbers.
But actually in this case,
since they're nondecreasing sequences,
if they don't tend to infinity,
then they never ... not only do they not tend to infinity,
but actually, in this case, none of the terms
at all will be infinity, either.
'cause if a nondecreasing sequence hits infinity it gets stuck there.
So, so you just get the
completely ordinary algebra of limits in R in that case.
So we can assume at least one of them tends to infinity.
But there is an- ... one more special case.
If either of the sequences tends to 0,
then it actually has to be constantly 0,
because they're non, they're non-negative numbers.
If either sequence tends to 0
then because it's nondecreasing and it's non-negative,
then it must actually be the constant sequence 0.
And then that's an easy special case.
The remaining case is where neither of them is constantly 0.
So they both leave 0 eventually,
and never get back anywhere near 0.
And at least one of them tends to infinity.
And that final special case you can
do with a sort of Comparison Test.
Something, you'd have to work out some sort of
comparison test or
Sandwich Theorem, or deal with it directly by M's and N's.
So the remaining case,
both sequences eventually leave 0,
and at least one of them tends to infinity.
... that's "leave 0" ...
So then you have to prove that x_n y_n tends to infinity.
It's probably best to go back to
the original definition, the one in terms of M's and N's.
Or you could prove separately some results about
constant multiples by ... non-in-,
non-infinite multiples, which would be,
that would require something like that,
you could prove a result about constant multiples separately,
and then use the Sandwich Theorem. That would also work.
And I'll let you fill in the details however you like on that one.
So this is absolutely crucial when it
comes to some of the theory.
We may not get to all the theory I want to get to.
But if you want to get through any of the integration theory
and so on, you will absolutely ...
it's absolutely essential that you
know that you can work with monotone increasing
products like this, because you need some sort of algebra of limits,
still, even though we've lost the,
we've lost the fully general algebra of limits for products.
But you need something still in order to do any sensible work at all.
So this is what we've got left.
Any questions about the algebra of limits?
I think that's the end of the algebra of limits section.
Would you like me to say anything more about any of the ...
any of the algebra of limits?
Ok! Well, the main thing, of course,
because when I don't give you the proofs,
that means the proofs are not examinable as bookwork,
and ... but, of course,
things can come up as unseen, or on assessed
coursework or something like that, and anything,
of course, any work you do in filling in
the details will help you to understand this stuff better.
So it's all worth filling in, if you can do it.
But if you want to just take my word for it,
and believe the results that I've stated, then that will be sufficient
for, for understanding the syllabus, even if
it doesn't get you as fluent as possible with all the material.
It'll still, you'll still have enough to be able
to follow the notes. Right, so let's
stop that recording there.