WEBVTT

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Welcome back to recitation.

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In this video I
want to talk to you

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about another test
for convergence

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we have for series,
that you haven't really

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spent any time looking
at this particular one.

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And it's pretty
helpful, and also will

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help us understand something
about Taylor series, which

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I'll do in another video.

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So this is the ratio test.

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And you'll understand the name
of this test, momentarily.

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So we're going to start
with a series-- sorry--

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we're going to start with a
series that we'll just say,

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we'll call each term a sub n.

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And I'm not going to
tell you where n starts,

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because it doesn't matter.

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It's really going to
only matter what's

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happening out at infinity.

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And to make things
simpler, we're

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going to let all the
terms be positive.

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OK?

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You, if they're
not positive, you

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can take the absolute
value of all the terms

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and still make some
conclusions in terms

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of absolute convergence.

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So that's just a little sidebar.

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But let's just deal with
all the terms positive,

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so we don't have to
worry about anything.

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And now, what does
the ratio test say?

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Well, the ratio test
says that first, we

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consider a certain ratio.

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The limit as n goes to infinity
of a sub n plus 1 over a sub n.

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OK?

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So we consider this limit.

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Well, that's your ratio.

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What is this doing?

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This is taking a term, and it's
dividing by the previous term.

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You do that for all
of the values for n,

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as n goes to infinity,
and you look at the limit,

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if it exists.

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OK?

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If the limit doesn't exist,
then you can't use this test.

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So sometimes that will happen.

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But if the limit exists,
you can use this test,

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and you say it equals L.

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And then you have the following
conclusions you can make.

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So here are the conclusions.

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There's three of them.

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So if L is less than 1,
then the series converges.

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OK?

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That's nice.

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That's good.

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If L is bigger than 1,
the series diverges.

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OK?

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That's another good thing.

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And then the last one
is, if L equals 1,

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you can't conclude anything.

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So I will try and
convince you of that fact

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with a few examples later.

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But let's look at this.

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So you look at the ratio, and if
the ratio is less than 1, then

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you actually can conclude
the series converges.

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And if the ratio
is bigger than 1,

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you can conclude that
the series diverges.

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So let me just give you
a little understanding

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of why this one
is true, and then

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the same kind of logic can
be used for this second one.

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So we'll try and understand
just at least a little bit

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why, when L is less than
1, the series converges.

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OK.

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So let me just
start writing here.

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So if L is less
than 1, then this

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means that we have that a
sub n plus 1 over a sub n

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as n goes to infinity is equal
to L, which is less than 1.

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Right?

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So we can pick something
between L and 1.

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We can pick a number
between 1 and 1,

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because L is
strictly less than 1.

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And so what I'm going to
do, is I'm going to say,

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I'm going to call this thing r.

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Some number between L and 1.

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OK?

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So what does that mean?

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That means that for large
n, we have a sub n plus 1

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over a sub n-- sorry, that
looks like I'm adding 1

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to the a sub n-- that's
a subscript-- a sub

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n plus 1 over a sub n is
less than some fixed r.

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So I'm picking a value
r between L and 1.

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And then this is
true for all large N.

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And when you're doing
math, sometimes people

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say, for n, you
know, bigger than

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or equal to some fixed value.

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Basically, if you go far
enough out in the sequence,

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then all the values bigger than
some fixed 1 have this ratio.

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OK?

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But let's get fancy with this.

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This we can rewrite as r to
the n plus 1 over r to the n.

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Right?

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This is just r.

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And now if I do a little
moving around, what do I see?

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I see that a sub n plus
1 over r to the n plus 1

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is less than a sub
n over r to the n.

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Now, this might be weird.

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What did I do?

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I just multiplied
through by the a sub n,

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and I divided by
the r sub n plus 1.

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I get this thing, and then
I see that this ratio,

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as n goes to infinity, the ratio
between a sub n and r to the n

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is decreasing.

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It's a decreasing, because
the next term, it's smaller.

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Right?

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And so the point is
that if I go far enough

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out, if I start, say,
past this n naught,

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if I go far enough
out, then I always have

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that a sub n is less than some
constant times r to the n.

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OK?

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So this means-- this
implies-- a sub n is always

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less than some constant
times r to the n.

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Right?

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And now, what do we do?

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We do our comparison test.

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OK?

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We do our comparison test.

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This is what's going to tell
us that the series converges.

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Now, again, this
isn't necessarily true

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all the way through the
series, but it's true

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when you're far enough
out, after this n naught.

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And if a series
converges at the end,

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the beginning is
just a finite sum.

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So we don't have to
worry about what's

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going on at the beginning.

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So again, we're at this place.

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We want to know what happens
to the sum of a sub n,

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of all the terms a sub n.

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Well, we know that's going to
be less than k times the sum

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r to the n.

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Right?

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Because each a sub n is less
than some constant times r

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to the n.

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Now why does this converge
What do we know about r?

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r we chose, we said
it's between L and 1.

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In particular, it's less than 1.

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This is a geometric series.

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Geometric series, when r is less
than 1, we know it converges.

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And so this one converges,
so then this one converges.

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So that's the logic behind it.

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We're going to now-- you know,
we're going to now use it.

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But I want to point
out that if you

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liked that, you can
come back over here,

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and you can do the same
kind of reasoning for why,

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if L is bigger than 1, a
sub n, the series, the sum

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of the a sub n's diverges.

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And that's going
to come down to,

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now you choose an r
that's between L and 1,

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but it has to be bigger than 1.

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OK?

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And then you can you
can look at that.

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Or maybe r has to be bigger
than L. I didn't even

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work that one out all the way.

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But you can do it.

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You put an r in there somewhere.

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And the same kind of logic,
because the r will be bigger

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than one, you're going
to get to a place

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where probably the
inequality sign is going

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to go the opposite way, right?

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And then you'll have a
series that diverges,

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and the other one will
be bigger than that one.

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And so that's how the
logic is going to work.

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So you have to figure
out where the r goes,

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but I guarantee you'll
want the r bigger than 1.

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And then you can, you'll have
to have the inequality signs

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be opposite what they are here.

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OK?

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You'll see, they're going
to be opposite there.

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OK.

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Now let's get some examples.

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Example 1.

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Let's look at some
that we know, and then

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let's look at some
that we don't know.

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OK?

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So let's look for example
first at 1 over n.

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Alright?

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Let's use the ratio
test on 1 over n.

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Maybe this seems funny, 'cause
what do we know about it?

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We know it diverges, right?

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But let's check,
if this tells us.

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The limit of n goes
to infinity of-- well,

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what's the n plus
first term of this?

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It's going to be
1 over n plus 1.

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And what's the nth term of this?

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It's going to be 1 over n.

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And so we get,
it's the limit as n

00:07:55.650 --> 00:08:03.070
goes to infinity of n over
n plus 1, and that equals 1.

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Hmm.

00:08:04.430 --> 00:08:05.800
So this one didn't work.

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This one didn't
tell us anything.

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And, OK.

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But we know this one diverges.

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So we know that-- this makes us
think, well, maybe when l is 1,

00:08:15.150 --> 00:08:16.280
then we know it diverges.

00:08:16.280 --> 00:08:20.470
But just to make sure we
don't make that conclusion,

00:08:20.470 --> 00:08:23.580
we don't draw that conclusion,
let's look at 1 over n squared.

00:08:23.580 --> 00:08:26.610
And what are the terms there?
a sub n plus 1 and a sub n.

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The limit as n goes to infinity.

00:08:29.280 --> 00:08:33.590
Well, the n plus first term
is going to be 1 over n

00:08:33.590 --> 00:08:36.370
plus 1 quantity squared,
and the nth term

00:08:36.370 --> 00:08:38.930
is going to be 1
over n squared, which

00:08:38.930 --> 00:08:44.530
is going to be the limit as n
goes to infinity of n squared

00:08:44.530 --> 00:08:49.040
over n plus 1 quantity squared,
which is also equal to 1.

00:08:49.040 --> 00:08:50.900
And what do we know
about this one?

00:08:50.900 --> 00:08:52.380
This one converges.

00:08:52.380 --> 00:08:55.510
So this one gave us
the L is equal to 1,

00:08:55.510 --> 00:08:56.810
but we know this one diverges.

00:08:56.810 --> 00:09:00.020
And this one gave us L is equal
to 1, and we know it converges.

00:09:00.020 --> 00:09:03.090
So we know that when L equals
1, we really cannot conclude

00:09:03.090 --> 00:09:04.565
convergence or divergence.

00:09:04.565 --> 00:09:05.065
OK?

00:09:05.065 --> 00:09:08.000
L equals 1 doesn't let
us draw any conclusions.

00:09:08.000 --> 00:09:10.300
But now let's see
something where, you know,

00:09:10.300 --> 00:09:11.800
we can draw a
conclusion, because it

00:09:11.800 --> 00:09:15.470
would be no fun if this
test never told us anything.

00:09:15.470 --> 00:09:18.140
It probably wouldn't
be a test, then.

00:09:18.140 --> 00:09:19.520
So let's try this one.

00:09:19.520 --> 00:09:26.130
Let's try 4 to the n
over n times 3 to n.

00:09:26.130 --> 00:09:28.080
Let's see what that one does.

00:09:28.080 --> 00:09:28.580
All right?

00:09:28.580 --> 00:09:30.580
So let's see.

00:09:30.580 --> 00:09:33.480
What is, we need the limit
as n goes to infinity.

00:09:33.480 --> 00:09:37.600
We need the n plus first term,
so let's plug in n plus 1

00:09:37.600 --> 00:09:39.250
for all of these.

00:09:39.250 --> 00:09:42.170
And I'm actually going to
do a little trick here,

00:09:42.170 --> 00:09:43.970
and I'll explain it as I go.

00:09:43.970 --> 00:09:49.310
4 to the n plus 1 over n
plus 1 3 to the n plus 1.

00:09:49.310 --> 00:09:52.697
I'm going to multiply
by 1 over a sub n.

00:09:52.697 --> 00:09:54.530
Because sometimes that's
a lot easier to do.

00:09:54.530 --> 00:09:56.530
I could have done it on
these other ones, maybe,

00:09:56.530 --> 00:09:58.420
but now I'm going to
do it on this one.

00:09:58.420 --> 00:10:00.940
So this is actually
what the a sub

00:10:00.940 --> 00:10:02.350
n is going to look like, right?

00:10:02.350 --> 00:10:05.240
I had to put in n plus
1 to get a sub n plus 1.

00:10:05.240 --> 00:10:06.640
This is a sub n.

00:10:06.640 --> 00:10:09.020
I'm going to write
down 1 over a sub n,

00:10:09.020 --> 00:10:12.770
and that's going to give me
n times 3 to the n over 4

00:10:12.770 --> 00:10:14.250
to the n.

00:10:14.250 --> 00:10:16.580
And now let's start simplifying.

00:10:16.580 --> 00:10:20.030
I have 3 to the n over
3 to the n plus 1.

00:10:20.030 --> 00:10:22.270
I'm left with just a 3 there.

00:10:22.270 --> 00:10:25.580
4 to the n and 4
to the n plus 1.

00:10:25.580 --> 00:10:27.311
I'm left with just a 4 there.

00:10:27.311 --> 00:10:29.810
And now the limit as n goes to
infinity-- let's just rewrite

00:10:29.810 --> 00:10:30.910
it so I know what it is.

00:10:34.850 --> 00:10:35.350
Let's see.

00:10:35.350 --> 00:10:41.170
I have 4n over 3 times n plus 1.

00:10:41.170 --> 00:10:43.570
Well, the 4 and the 3 I
can actually just pull out.

00:10:43.570 --> 00:10:45.470
But what did I have
here? n over n plus 1?

00:10:45.470 --> 00:10:49.550
The limit of n goes to infinity
of that, that equals to 4/3.

00:10:49.550 --> 00:10:51.130
That's bigger than 1.

00:10:51.130 --> 00:10:54.955
So that actually diverges, OK?

00:11:00.586 --> 00:11:02.960
And I have one more example,
and I'm almost out of space.

00:11:02.960 --> 00:11:05.190
And let me actually come
over here and figure out

00:11:05.190 --> 00:11:06.430
what example I wanted.

00:11:06.430 --> 00:11:07.330
Ah.

00:11:07.330 --> 00:11:08.140
Sorry about that.

00:11:08.140 --> 00:11:09.770
I knew I had one more.

00:11:09.770 --> 00:11:12.450
OK.

00:11:12.450 --> 00:11:15.423
n to the tenth over 10 to the n.

00:11:15.423 --> 00:11:17.506
So it's kind of interesting
one, because you have,

00:11:17.506 --> 00:11:21.450
you have an exponential and
then you have a power of n.

00:11:21.450 --> 00:11:22.990
So let's look at this one.

00:11:22.990 --> 00:11:26.790
So we need to consider
limit as n goes to infinity.

00:11:26.790 --> 00:11:29.520
So I put in n plus 1 first.

00:11:29.520 --> 00:11:35.420
n plus 1 to the tenth
over 10 to the n plus 1.

00:11:35.420 --> 00:11:38.260
And then I'm going to
just, remember, do times

00:11:38.260 --> 00:11:39.750
1 over a sub n.

00:11:39.750 --> 00:11:43.450
So I'm going to have 10 to
the n over n to the tenth.

00:11:43.450 --> 00:11:46.530
So again, I took
a sub n, I did 1

00:11:46.530 --> 00:11:48.770
over that, that's
just the reciprocal.

00:11:48.770 --> 00:11:51.680
And now let's start dividing,
if I'm allowed to divide.

00:11:51.680 --> 00:11:53.090
Yeah, I've got 10
to the n there,

00:11:53.090 --> 00:11:55.420
and 10 to the n plus 1
there, so that gives me

00:11:55.420 --> 00:11:57.690
a single 10 in the denominator.

00:11:57.690 --> 00:12:00.640
And so now I really
have the limit

00:12:00.640 --> 00:12:07.200
as n goes to infinity of
1 over 10 times n plus 1

00:12:07.200 --> 00:12:11.734
to the tenth over
n to the tenth.

00:12:11.734 --> 00:12:13.900
Well, that's equal to-- n
plus 1 to the tenth over n

00:12:13.900 --> 00:12:14.441
to the tenth,

00:12:14.441 --> 00:12:17.060
you might start to get nervous
and think, "Oh my gosh!

00:12:17.060 --> 00:12:18.560
These powers are
getting really big!

00:12:18.560 --> 00:12:20.790
It might make a difference
that that 1 is there."

00:12:20.790 --> 00:12:22.570
It doesn't make a difference
that that 1 is there,

00:12:22.570 --> 00:12:24.236
because if you actually
expand this out,

00:12:24.236 --> 00:12:26.340
the leading term is
just n to the tenth.

00:12:26.340 --> 00:12:30.975
And we know that the highest
order, or the highest degree

00:12:30.975 --> 00:12:32.470
is going to win
out, so it's going

00:12:32.470 --> 00:12:34.636
to be n to the tenth over
n to the tenth is how it's

00:12:34.636 --> 00:12:35.920
going to behave in the limit.

00:12:35.920 --> 00:12:39.000
So this part's just going
to go to 1, so I get 1/10.

00:12:39.000 --> 00:12:40.321
Oh, that's less than 1!

00:12:40.321 --> 00:12:40.820
Yay!

00:12:40.820 --> 00:12:44.350
So this series converges.

00:12:47.621 --> 00:12:48.120
OK?

00:12:48.120 --> 00:12:50.170
So we had one that diverges,
one that converges,

00:12:50.170 --> 00:12:53.620
and a few where we couldn't
get conclusions by this test.

00:12:53.620 --> 00:12:55.800
And one point I want
to make about this,

00:12:55.800 --> 00:12:59.440
is that in some cases, you
have the integral test already,

00:12:59.440 --> 00:13:01.920
and sometimes that's
easy and that helps you.

00:13:01.920 --> 00:13:03.580
In the case of
these examples where

00:13:03.580 --> 00:13:05.390
we couldn't tell where
l was equal to 1,

00:13:05.390 --> 00:13:07.630
the integral test is going
to tell you something.

00:13:07.630 --> 00:13:09.470
But in this case, it's
a little bit harder

00:13:09.470 --> 00:13:13.540
to deal with this,
as an integral test.

00:13:13.540 --> 00:13:16.190
You can still do it, but
it's a little bit harder.

00:13:16.190 --> 00:13:17.870
And so this, maybe,
is a little bit

00:13:17.870 --> 00:13:21.020
quicker way to deal with these
types of, types of problems.

00:13:21.020 --> 00:13:22.880
And the big thing
we're going to do,

00:13:22.880 --> 00:13:24.505
is in the next video
on the ratio test,

00:13:24.505 --> 00:13:25.880
I'm going to show
you how you can

00:13:25.880 --> 00:13:28.050
use this to determine
the radius of convergence

00:13:28.050 --> 00:13:29.480
for these Taylor series.

00:13:29.480 --> 00:13:31.830
So that's actually going
to be kind of exciting,

00:13:31.830 --> 00:13:34.048
and that'll be in
our next video.