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Welcome back to recitation.

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In this video I want to talk to
you about another test for

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convergence we have for series,
that you haven't

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really spent any time looking
at this particular one.

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And it's pretty helpful, and
also will help us understand

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something about Taylor
series, which I'll

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do in another video.

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So this is the ratio test. And
you'll understand the name of

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this test, momentarily.

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

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

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each term a sub n.

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

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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 going to let all the

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terms be positive.

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OK?

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You, if they're not positive,
you can take the absolute

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value of all the terms and still
make some conclusions in

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terms 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, so we

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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

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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

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the previous term.

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You do that for all of the
values for n, as n goes to

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infinity, and you look at
the limit, 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, and you say

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it equals l.

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And then you have the following

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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 with a few

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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,

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then you actually can conclude
the series converges.

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And if the ratio is bigger than
1, you can conclude that

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the series diverges.

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So let me just give you a little
understanding of why

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this one is true, and then the
same kind of logic can be used

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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 serious 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 means that we have that a

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sub n plus 1 over a sub n as n
goes to infinity is equal to

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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, because I is strictly

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less than 1.

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

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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 over a

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

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that's a subscript--

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a sub 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. And when you're doing

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math, sometimes people 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

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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 is less than

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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, and I divided by

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the r sub n plus 1.

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I get this thing, and then I see
that this ratio, as n goes

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to infinity, the ratio between
a sub n and r to the n is

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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 out, if I start,

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say, past this n naught, if I
go far enough out, then I

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always have that a sub n is less
than some constant times

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r to the n.

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OK?

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So this means--

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this is implies--

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a sub n is always less
than some constant

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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. OK?

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We do our comparison test. This
is what's going to tell

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us that the series converges.

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Now, again, this isn't
necessarily true all the way

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through the series, but it's
true when you're far enough

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out, after this n naught.

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And if a series converges at the
end, the beginning is just

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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 of all

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the terms a sub n.

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

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sum r to the n.

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Right?

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Because each a sub n is less
than some constant

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times r 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

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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 liked that, you can

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come back over here, and you
can do the same kind of

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reasoning for why, if l is
bigger than 1, a sub n, the

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series, the sum of the
a sub n's diverges.

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And that's going to come down
to, now you choose an r that's

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between l and 1, 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.

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I didn't even 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 where probably the

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inequality sign is going to go
the opposite way, right?

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And then you'll have a series
that diverges, and the other

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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, but I

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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 let's look at

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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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And 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--

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well, 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 goes to infinity of n

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over n plus 1, and
that equals 1.

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Hm.

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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

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1, then we know it diverges.

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

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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 plus 1

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quantity squared, and the nth
term is going to be 1 over n

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squared, which is going to
be the limit as n goes to

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infinity of n squared over n
plus 1 quantity squared, which

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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 one, and but we know

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this one diverges.

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And this one gave us
l is equal to 1,

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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, we can draw a

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conclusion, because it would be
no fun if this test never

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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
of n goes to infinity.

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

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all of these.

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

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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, but now I'm

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going to do it on this one.

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So this is actually what
the a sub n is going

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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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I'm going to write down 1 over
a sub n, and that's going to

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give me n times 3 to the
n over 4 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.

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I'm left with just a 4 there.

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00:10:27,320 --> 00:10:28,970
And now the limit as
n goes to infinity.

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Let's just rewrite it so
I know what it is.

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00:10:30,910 --> 00:10:34,940

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Let's see.

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I have 4n over 3
times n plus 1.

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Well, the 4 and the 3 I can
actually just pull out.

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

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The limit of n goes to
infinity of that,

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that equals to 4/3.

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That's bigger than 1.

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So that actually diverges, OK?

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00:10:54,955 --> 00:11:00,720

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00:11:00,720 --> 00:11:02,960
And I have one more example, and
I'm almost out of space.

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And let me actually come over
here and figure out what

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example I wanted.

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Ah.

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

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I knew I had one more.

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OK.

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00:11:12,450 --> 00:11:15,960
n to the tenth over
10 to the n.

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So it's kind of interesting
one, because you have, you

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have an exponential and then
you have a power of n.

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So let's look at this one.

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So we need to consider limit
as n goes to infinity.

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So I put in n plus 1 first. n
plus 1 to the tenth over 10 to

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the n plus 1.

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And then I'm going to just,
remember, do times

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

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So I'm going to have 10 to the
n over n to the tenth.

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So again, I took a sub n, I did
1 over that, that's just

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00:11:47,480 --> 00:11:48,770
the reciprocal.

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

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00:11:51,680 --> 00:11:54,810
Yeah, I've got 10 the n there,
and 10 to the n plus 1 there,

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00:11:54,810 --> 00:11:57,690
so that gives me a single
10 in the denominator.

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00:11:57,690 --> 00:12:03,470
And so now I really have the
limit as n goes to infinity of

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00:12:03,470 --> 00:12:11,820
1 over 10 times n plus 1 to the
tenth over n to the tenth.

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00:12:11,820 --> 00:12:13,560
Well, that's equal to n
plus 1 to the tenth

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00:12:13,560 --> 00:12:14,330
over n to the tenth.

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00:12:14,330 --> 00:12:17,140
You might start to get nervous
and think, "Oh my gosh!

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00:12:17,140 --> 00:12:18,560
These powers are getting
really big!

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00:12:18,560 --> 00:12:20,970
It might make a difference
that that 1 is there." It

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00:12:20,970 --> 00:12:22,820
doesn't make a difference that
that 1 is there, because if

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00:12:22,820 --> 00:12:24,960
you actually expand this out,
the leading term is

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00:12:24,960 --> 00:12:26,340
just n to the tenth.

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

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00:12:30,975 --> 00:12:33,360
is going to win out, so it's
going to be n to the tenth

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00:12:33,360 --> 00:12:34,780
over n to the tenth is
how it's going to

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00:12:34,780 --> 00:12:35,920
behave in the limit.

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

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00:12:39,000 --> 00:12:40,480
Oh, that's less than 1!

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00:12:40,480 --> 00:12:40,820
Yay!

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

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00:12:44,350 --> 00:12:47,730

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00:12:47,730 --> 00:12:48,120
OK?

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00:12:48,120 --> 00:12:50,610
So we had one that diverges, one
that converges, and a few

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00:12:50,610 --> 00:12:54,070
where we couldn't get
conclusions by this test. And

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00:12:54,070 --> 00:12:57,780
one point I want to make about
this, is that in some cases,

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00:12:57,780 --> 00:13:00,790
you have the integral test
already, and sometimes that's

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00:13:00,790 --> 00:13:01,920
easy and that helps you.

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00:13:01,920 --> 00:13:04,730
In the case of these examples
where we couldn't tell where l

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00:13:04,730 --> 00:13:06,110
was equal to 1, the
integral test is

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00:13:06,110 --> 00:13:07,630
going to tell you something.

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00:13:07,630 --> 00:13:11,430
But in this case, it's a little
bit harder to deal with

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00:13:11,430 --> 00:13:15,120
this as an integral test. You
can still do it, but it's a

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00:13:15,120 --> 00:13:16,190
little bit harder.

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00:13:16,190 --> 00:13:18,730
And so this, maybe, is a little
bit quicker way to deal

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00:13:18,730 --> 00:13:21,020
with these types of,
types of problems.

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00:13:21,020 --> 00:13:24,070
And the big thing we're going to
do, is in the next video on

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00:13:24,070 --> 00:13:26,340
the ratio test, I'm going to
show you how you can use this

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00:13:26,340 --> 00:13:28,190
to determine the radius
of convergence for

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00:13:28,190 --> 00:13:29,480
these Taylor series.

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00:13:29,480 --> 00:13:32,270
So that's actually going to be
kind of exciting, and that'll

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00:13:32,270 --> 00:13:33,520
be in our next video.

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