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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

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goes to infinity of n over
n plus 1, and that equals 1.

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Hmm.

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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,

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then we know it diverges.

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But just to make sure we
don't make that conclusion,

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we don't draw that conclusion,
let's look at 1 over n squared.

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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.

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Well, the n plus first term
is going to be 1 over n

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plus 1 quantity squared,
and the nth term

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is going to be 1
over n squared, which

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is going to be the limit as n
goes to infinity of n squared

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over n plus 1 quantity squared,
which is also equal to 1.

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And what do we know
about this one?

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This one converges.

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So this one gave us
the L is equal to 1,

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but we know this one diverges.

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And this one gave us L is equal
to 1, and we know it converges.

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So we know that when L equals
1, we really cannot conclude

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convergence or divergence.

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OK?

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L equals 1 doesn't let
us draw any conclusions.

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But now let's see
something where, you know,

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we can draw a
conclusion, because it

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would be no fun if this
test never told us anything.

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It probably wouldn't
be a test, then.

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So let's try this one.

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Let's try 4 to the n
over n times 3 to n.

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Let's see what that one does.

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All right?

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So let's see.

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What is, we need the limit
as n goes to infinity.

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We need the n plus first term,
so let's plug in n plus 1

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for all of these.

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And I'm actually going to
do a little trick here,

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and I'll explain it as I go.

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4 to the n plus 1 over n
plus 1 3 to the n plus 1.

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I'm going to multiply
by 1 over a sub n.

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Because sometimes that's
a lot easier to do.

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I could have done it on
these other ones, maybe,

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but now I'm going to
do it on this one.

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So this is actually
what the a sub

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n is going to look like, right?

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I had to put in n plus
1 to get a sub n plus 1.

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This is a sub n.

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00:10:06,640 --> 00:10:09,020
I'm going to write
down 1 over a sub n,

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and that's going to give me
n times 3 to the n over 4

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to the n.

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And now let's start simplifying.

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I have 3 to the n over
3 to the n plus 1.

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I'm left with just a 3 there.

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4 to the n and 4
to the n plus 1.

253
00:10:25,580 --> 00:10:27,311
I'm left with just a 4 there.

254
00:10:27,311 --> 00:10:29,810
And now the limit as n goes to
infinity-- let's just rewrite

255
00:10:29,810 --> 00:10:30,910
it so I know what it is.

256
00:10:30,910 --> 00:10:34,850


257
00:10:34,850 --> 00:10:35,350
Let's see.

258
00:10:35,350 --> 00:10:41,170
I have 4n over 3 times n plus 1.

259
00:10:41,170 --> 00:10:43,570
Well, the 4 and the 3 I
can actually just pull out.

260
00:10:43,570 --> 00:10:45,470
But what did I have
here? n over n plus 1?

261
00:10:45,470 --> 00:10:49,550
The limit of n goes to infinity
of that, that equals to 4/3.

262
00:10:49,550 --> 00:10:51,130
That's bigger than 1.

263
00:10:51,130 --> 00:10:54,955
So that actually diverges, OK?

264
00:10:54,955 --> 00:11:00,586


265
00:11:00,586 --> 00:11:02,960
And I have one more example,
and I'm almost out of space.

266
00:11:02,960 --> 00:11:05,190
And let me actually come
over here and figure out

267
00:11:05,190 --> 00:11:06,430
what example I wanted.

268
00:11:06,430 --> 00:11:07,330
Ah.

269
00:11:07,330 --> 00:11:08,140
Sorry about that.

270
00:11:08,140 --> 00:11:09,770
I knew I had one more.

271
00:11:09,770 --> 00:11:12,450
OK.

272
00:11:12,450 --> 00:11:15,423
n to the tenth over 10 to the n.

273
00:11:15,423 --> 00:11:17,506
So it's kind of interesting
one, because you have,

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00:11:17,506 --> 00:11:21,450
you have an exponential and
then you have a power of n.

275
00:11:21,450 --> 00:11:22,990
So let's look at this one.

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00:11:22,990 --> 00:11:26,790
So we need to consider
limit as n goes to infinity.

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00:11:26,790 --> 00:11:29,520
So I put in n plus 1 first.

278
00:11:29,520 --> 00:11:35,420
n plus 1 to the tenth
over 10 to the n plus 1.

279
00:11:35,420 --> 00:11:38,260
And then I'm going to
just, remember, do times

280
00:11:38,260 --> 00:11:39,750
1 over a sub n.

281
00:11:39,750 --> 00:11:43,450
So I'm going to have 10 to
the n over n to the tenth.

282
00:11:43,450 --> 00:11:46,530
So again, I took
a sub n, I did 1

283
00:11:46,530 --> 00:11:48,770
over that, that's
just the reciprocal.

284
00:11:48,770 --> 00:11:51,680
And now let's start dividing,
if I'm allowed to divide.

285
00:11:51,680 --> 00:11:53,090
Yeah, I've got 10
to the n there,

286
00:11:53,090 --> 00:11:55,420
and 10 to the n plus 1
there, so that gives me

287
00:11:55,420 --> 00:11:57,690
a single 10 in the denominator.

288
00:11:57,690 --> 00:12:00,640
And so now I really
have the limit

289
00:12:00,640 --> 00:12:07,200
as n goes to infinity of
1 over 10 times n plus 1

290
00:12:07,200 --> 00:12:11,734
to the tenth over
n to the tenth.

291
00:12:11,734 --> 00:12:13,900
Well, that's equal to-- n
plus 1 to the tenth over n

292
00:12:13,900 --> 00:12:14,441
to the tenth,

293
00:12:14,441 --> 00:12:17,060
you might start to get nervous
and think, "Oh my gosh!

294
00:12:17,060 --> 00:12:18,560
These powers are
getting really big!

295
00:12:18,560 --> 00:12:20,790
It might make a difference
that that 1 is there."

296
00:12:20,790 --> 00:12:22,570
It doesn't make a difference
that that 1 is there,

297
00:12:22,570 --> 00:12:24,236
because if you actually
expand this out,

298
00:12:24,236 --> 00:12:26,340
the leading term is
just n to the tenth.

299
00:12:26,340 --> 00:12:30,975
And we know that the highest
order, or the highest degree

300
00:12:30,975 --> 00:12:32,470
is going to win
out, so it's going

301
00:12:32,470 --> 00:12:34,636
to be n to the tenth over
n to the tenth is how it's

302
00:12:34,636 --> 00:12:35,920
going to behave in the limit.

303
00:12:35,920 --> 00:12:39,000
So this part's just going
to go to 1, so I get 1/10.

304
00:12:39,000 --> 00:12:40,321
Oh, that's less than 1!

305
00:12:40,321 --> 00:12:40,820
Yay!

306
00:12:40,820 --> 00:12:44,350
So this series converges.

307
00:12:44,350 --> 00:12:47,621


308
00:12:47,621 --> 00:12:48,120
OK?

309
00:12:48,120 --> 00:12:50,170
So we had one that diverges,
one that converges,

310
00:12:50,170 --> 00:12:53,620
and a few where we couldn't
get conclusions by this test.

311
00:12:53,620 --> 00:12:55,800
And one point I want
to make about this,

312
00:12:55,800 --> 00:12:59,440
is that in some cases, you
have the integral test already,

313
00:12:59,440 --> 00:13:01,920
and sometimes that's
easy and that helps you.

314
00:13:01,920 --> 00:13:03,580
In the case of
these examples where

315
00:13:03,580 --> 00:13:05,390
we couldn't tell where
l was equal to 1,

316
00:13:05,390 --> 00:13:07,630
the integral test is going
to tell you something.

317
00:13:07,630 --> 00:13:09,470
But in this case, it's
a little bit harder

318
00:13:09,470 --> 00:13:13,540
to deal with this,
as an integral test.

319
00:13:13,540 --> 00:13:16,190
You can still do it, but
it's a little bit harder.

320
00:13:16,190 --> 00:13:17,870
And so this, maybe,
is a little bit

321
00:13:17,870 --> 00:13:21,020
quicker way to deal with these
types of, types of problems.

322
00:13:21,020 --> 00:13:22,880
And the big thing
we're going to do,

323
00:13:22,880 --> 00:13:24,505
is in the next video
on the ratio test,

324
00:13:24,505 --> 00:13:25,880
I'm going to show
you how you can

325
00:13:25,880 --> 00:13:28,050
use this to determine
the radius of convergence

326
00:13:28,050 --> 00:13:29,480
for these Taylor series.

327
00:13:29,480 --> 00:13:31,830
So that's actually going
to be kind of exciting,

328
00:13:31,830 --> 00:13:34,048
and that'll be in
our next video.

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00:13:34,048 --> 00:13:34,548