Welcome back to recitation.
In this video I
want to talk to you
about another test
for convergence
we have for series,
that you haven't really
spent any time looking
at this particular one.
And it's pretty
helpful, and also will
help us understand something
about Taylor series, which
I'll do in another video.
So this is the ratio test.
And you'll understand the name
of this test, momentarily.
So we're going to start
with a series-- sorry--
we're going to start with a
series that we'll just say,
we'll call each term a sub n.
And I'm not going to
tell you where n starts,
because it doesn't matter.
It's really going to
only matter what's
happening out at infinity.
And to make things
simpler, we're
going to let all the
terms be positive.
OK?
You, if they're
not positive, you
can take the absolute
value of all the terms
and still make some
conclusions in terms
of absolute convergence.
So that's just a little sidebar.
But let's just deal with
all the terms positive,
so we don't have to
worry about anything.
And now, what does
the ratio test say?
Well, the ratio test
says that first, we
consider a certain ratio.
The limit as n goes to infinity
of a sub n plus 1 over a sub n.
OK?
So we consider this limit.
Well, that's your ratio.
What is this doing?
This is taking a term, and it's
dividing by the previous term.
You do that for all
of the values for n,
as n goes to infinity,
and you look at the limit,
if it exists.
OK?
If the limit doesn't exist,
then you can't use this test.
So sometimes that will happen.
But if the limit exists,
you can use this test,
and you say it equals L.
And then you have the following
conclusions you can make.
So here are the conclusions.
There's three of them.
So if L is less than 1,
then the series converges.
OK?
That's nice.
That's good.
If L is bigger than 1,
the series diverges.
OK?
That's another good thing.
And then the last one
is, if L equals 1,
you can't conclude anything.
So I will try and
convince you of that fact
with a few examples later.
But let's look at this.
So you look at the ratio, and if
the ratio is less than 1, then
you actually can conclude
the series converges.
And if the ratio
is bigger than 1,
you can conclude that
the series diverges.
So let me just give you
a little understanding
of why this one
is true, and then
the same kind of logic can
be used for this second one.
So we'll try and understand
just at least a little bit
why, when L is less than
1, the series converges.
OK.
So let me just
start writing here.
So if L is less
than 1, then this
means that we have that a
sub n plus 1 over a sub n
as n goes to infinity is equal
to L, which is less than 1.
Right?
So we can pick something
between L and 1.
We can pick a number
between 1 and 1,
because L is
strictly less than 1.
And so what I'm going to
do, is I'm going to say,
I'm going to call this thing r.
Some number between L and 1.
OK?
So what does that mean?
That means that for large
n, we have a sub n plus 1
over a sub n-- sorry, that
looks like I'm adding 1
to the a sub n-- that's
a subscript-- a sub
n plus 1 over a sub n is
less than some fixed r.
So I'm picking a value
r between L and 1.
And then this is
true for all large N.
And when you're doing
math, sometimes people
say, for n, you
know, bigger than
or equal to some fixed value.
Basically, if you go far
enough out in the sequence,
then all the values bigger than
some fixed 1 have this ratio.
OK?
But let's get fancy with this.
This we can rewrite as r to
the n plus 1 over r to the n.
Right?
This is just r.
And now if I do a little
moving around, what do I see?
I see that a sub n plus
1 over r to the n plus 1
is less than a sub
n over r to the n.
Now, this might be weird.
What did I do?
I just multiplied
through by the a sub n,
and I divided by
the r sub n plus 1.
I get this thing, and then
I see that this ratio,
as n goes to infinity, the ratio
between a sub n and r to the n
is decreasing.
It's a decreasing, because
the next term, it's smaller.
Right?
And so the point is
that if I go far enough
out, if I start, say,
past this n naught,
if I go far enough
out, then I always have
that a sub n is less than some
constant times r to the n.
OK?
So this means-- this
implies-- a sub n is always
less than some constant
times r to the n.
Right?
And now, what do we do?
We do our comparison test.
OK?
We do our comparison test.
This is what's going to tell
us that the series converges.
Now, again, this
isn't necessarily true
all the way through the
series, but it's true
when you're far enough
out, after this n naught.
And if a series
converges at the end,
the beginning is
just a finite sum.
So we don't have to
worry about what's
going on at the beginning.
So again, we're at this place.
We want to know what happens
to the sum of a sub n,
of all the terms a sub n.
Well, we know that's going to
be less than k times the sum
r to the n.
Right?
Because each a sub n is less
than some constant times r
to the n.
Now why does this converge
What do we know about r?
r we chose, we said
it's between L and 1.
In particular, it's less than 1.
This is a geometric series.
Geometric series, when r is less
than 1, we know it converges.
And so this one converges,
so then this one converges.
So that's the logic behind it.
We're going to now-- you know,
we're going to now use it.
But I want to point
out that if you
liked that, you can
come back over here,
and you can do the same
kind of reasoning for why,
if L is bigger than 1, a
sub n, the series, the sum
of the a sub n's diverges.
And that's going
to come down to,
now you choose an r
that's between L and 1,
but it has to be bigger than 1.
OK?
And then you can you
can look at that.
Or maybe r has to be bigger
than L. I didn't even
work that one out all the way.
But you can do it.
You put an r in there somewhere.
And the same kind of logic,
because the r will be bigger
than one, you're going
to get to a place
where probably the
inequality sign is going
to go the opposite way, right?
And then you'll have a
series that diverges,
and the other one will
be bigger than that one.
And so that's how the
logic is going to work.
So you have to figure
out where the r goes,
but I guarantee you'll
want the r bigger than 1.
And then you can, you'll have
to have the inequality signs
be opposite what they are here.
OK?
You'll see, they're going
to be opposite there.
OK.
Now let's get some examples.
Example 1.
Let's look at some
that we know, and then
let's look at some
that we don't know.
OK?
So let's look for example
first at 1 over n.
Alright?
Let's use the ratio
test on 1 over n.
Maybe this seems funny, 'cause
what do we know about it?
We know it diverges, right?
But let's check,
if this tells us.
The limit of n goes
to infinity of-- well,
what's the n plus
first term of this?
It's going to be
1 over n plus 1.
And what's the nth term of this?
It's going to be 1 over n.
And so we get,
it's the limit as n
goes to infinity of n over
n plus 1, and that equals 1.
Hmm.
So this one didn't work.
This one didn't
tell us anything.
And, OK.
But we know this one diverges.
So we know that-- this makes us
think, well, maybe when l is 1,
then we know it diverges.
But just to make sure we
don't make that conclusion,
we don't draw that conclusion,
let's look at 1 over n squared.
And what are the terms there?
a sub n plus 1 and a sub n.
The limit as n goes to infinity.
Well, the n plus first term
is going to be 1 over n
plus 1 quantity squared,
and the nth term
is going to be 1
over n squared, which
is going to be the limit as n
goes to infinity of n squared
over n plus 1 quantity squared,
which is also equal to 1.
And what do we know
about this one?
This one converges.
So this one gave us
the L is equal to 1,
but we know this one diverges.
And this one gave us L is equal
to 1, and we know it converges.
So we know that when L equals
1, we really cannot conclude
convergence or divergence.
OK?
L equals 1 doesn't let
us draw any conclusions.
But now let's see
something where, you know,
we can draw a
conclusion, because it
would be no fun if this
test never told us anything.
It probably wouldn't
be a test, then.
So let's try this one.
Let's try 4 to the n
over n times 3 to n.
Let's see what that one does.
All right?
So let's see.
What is, we need the limit
as n goes to infinity.
We need the n plus first term,
so let's plug in n plus 1
for all of these.
And I'm actually going to
do a little trick here,
and I'll explain it as I go.
4 to the n plus 1 over n
plus 1 3 to the n plus 1.
I'm going to multiply
by 1 over a sub n.
Because sometimes that's
a lot easier to do.
I could have done it on
these other ones, maybe,
but now I'm going to
do it on this one.
So this is actually
what the a sub
n is going to look like, right?
I had to put in n plus
1 to get a sub n plus 1.
This is a sub n.
I'm going to write
down 1 over a sub n,
and that's going to give me
n times 3 to the n over 4
to the n.
And now let's start simplifying.
I have 3 to the n over
3 to the n plus 1.
I'm left with just a 3 there.
4 to the n and 4
to the n plus 1.
I'm left with just a 4 there.
And now the limit as n goes to
infinity-- let's just rewrite
it so I know what it is.
Let's see.
I have 4n over 3 times n plus 1.
Well, the 4 and the 3 I
can actually just pull out.
But what did I have
here? n over n plus 1?
The limit of n goes to infinity
of that, that equals to 4/3.
That's bigger than 1.
So that actually diverges, OK?
And I have one more example,
and I'm almost out of space.
And let me actually come
over here and figure out
what example I wanted.
Ah.
Sorry about that.
I knew I had one more.
OK.
n to the tenth over 10 to the n.
So it's kind of interesting
one, because you have,
you have an exponential and
then you have a power of n.
So let's look at this one.
So we need to consider
limit as n goes to infinity.
So I put in n plus 1 first.
n plus 1 to the tenth
over 10 to the n plus 1.
And then I'm going to
just, remember, do times
1 over a sub n.
So I'm going to have 10 to
the n over n to the tenth.
So again, I took
a sub n, I did 1
over that, that's
just the reciprocal.
And now let's start dividing,
if I'm allowed to divide.
Yeah, I've got 10
to the n there,
and 10 to the n plus 1
there, so that gives me
a single 10 in the denominator.
And so now I really
have the limit
as n goes to infinity of
1 over 10 times n plus 1
to the tenth over
n to the tenth.
Well, that's equal to-- n
plus 1 to the tenth over n
to the tenth,
you might start to get nervous
and think, "Oh my gosh!
These powers are
getting really big!
It might make a difference
that that 1 is there."
It doesn't make a difference
that that 1 is there,
because if you actually
expand this out,
the leading term is
just n to the tenth.
And we know that the highest
order, or the highest degree
is going to win
out, so it's going
to be n to the tenth over
n to the tenth is how it's
going to behave in the limit.
So this part's just going
to go to 1, so I get 1/10.
Oh, that's less than 1!
Yay!
So this series converges.
OK?
So we had one that diverges,
one that converges,
and a few where we couldn't
get conclusions by this test.
And one point I want
to make about this,
is that in some cases, you
have the integral test already,
and sometimes that's
easy and that helps you.
In the case of
these examples where
we couldn't tell where
l was equal to 1,
the integral test is going
to tell you something.
But in this case, it's
a little bit harder
to deal with this,
as an integral test.
You can still do it, but
it's a little bit harder.
And so this, maybe,
is a little bit
quicker way to deal with these
types of, types of problems.
And the big thing
we're going to do,
is in the next video
on the ratio test,
I'm going to show
you how you can
use this to determine
the radius of convergence
for these Taylor series.
So that's actually going
to be kind of exciting,
and that'll be in
our next video.
