Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

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Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Convergence Tests

In this third module, we study various convergence tests to determine whether or not a series converges: in particular, we will consider the ratio test, the root test, and the integral test.

- Jim Fowler, PhDProfessor

Mathematics

The ratio test is awesome.

What test should I apply?

Well, for this series, the ratio test will work wonderfully.

I can really tell that the ratio test is just going to be great for this.

Because I've got these factorials and these powers, so

I can expect a lot of cancellation to happen.

Let's compute the limit of the ratio of neighboring terms.

So I'll write a sub n is n factorial over n

to the n, and I'm trying to calculate the limit, as n

approaches infinity of a sub n plus one over a sub n.

And that's the limit as n approaches infinity.

What's a sub n plus one?

I just gotta replace these n's with n plus one.

That's n plus one factorial divided by n plus one to the n plus oneth power divided

by what's a sub n.

Well, that's just n factorial over n to the nth power.

That can be simplified.

First of all, I've got a fraction with fractions in the

numerator and denominator, so I can clean that up a bit.

This is the limit as n approaches infinity.

Of n plus one factorial times n to the n, divided by n factorial

times n plus one to the n plus oneth power.

Now what can I do?

Well, I've got an n plus one factorial in numerator, and an n factorial in

the denominator, so this n factorial cancels everything

except for the n plus one term here.

So this is the limit as n approaches infinity of just n plus one

times n to the n.

Divided by, so that n factorial's gone now, n plus one to the n plus oneth power.

Well I've got an n plus one on the numerator, and the power of n plus one in

the denominator, so I can use this to change

this n plus one in just an nth power.

So this is the limit as n approaches infinity of n to the n over n

plus one to the n.

And if you like, I can just rewrite this a bit too or analyze this limit.

If you really love l'HĂ´pital's rule, you could just apply l'HĂ´pital's rule.

I don't really like l'HĂ´pital's rule that much.

So instead, I'm just going to recall a useful fact.

In fact, this might've been how you define the number e.

The limit, as n approaches infinity, of one plus one over n to the nth power is e.

Now, how can I take this fact and say something about this limit?

Well, I could combine this into a single fraction.

So one plus one over n is n over n plus one

over n, which means the limit of n plus one over n.

To the nth power as n approaches infinity is e.

And now this looks a whole lot like this. And indeed all I have to do,

is use the fact that the limit of a reciprocal is the reciprocal of

the limit to conclude that the limit of n over n

plus one to the nth power is in fact one over e.

What does that imply about the original series?

Now, one over e is less than one, and that means that, according to the

ratio test, the given series converges. We can do even better.

Does the series, n goes from one to infinity of n factorial

divided by n over two, to the nth power, converge or diverge?

Yes, this series converges.

Let's see why.

Well here we go.

Let's set a sub n equal to n factorial over N over two to the n.

And my claim

is that the sum, n goes from one to infinity of a, sub n converges.

To justify this claim, I'm going to use the ratio test.

So, big L, which is the limit as n approaches

infinity of a sub n plus one, over a sub n.

Well in this case what is that?

That's the limit as n approaches infinity of

this with n replaced by n plus one.

It's n plus one factorial over n plus one, over two to the n plus one power.

Divided by a sub n which is N factorial over N over two to the nth power.

This is kind of a mess because I've got fractions, a numerator, and a denominator.

So I can simplify that, can rewrite that as the

limit N goes to infinity of N plus one factorial

times n over two to the N divided by N

factorial times N plus one over two to the N plus one power.

I've got an N plus one factorial divided by an N factorial.

Most of those terms cancel except for the N plus one.

So, I can rewrite that as just N plus one on the numerator.

Let me simplify this a bit too, or at least let's expand it out.

I can write this as n to the

n, divided by two to the n.

So it's n to the n, divided by two to the n.

And the denominator here, well the n factorial goes away but I can rewrite this

as n plus n to the n plus one power, divided by two to the n plus one.

Now I can keep simplifying this.

I've got an an n plus one in the numerator, an n plus one to the

n plus one power in the denominator, I can cancel one of those n plus ones in

the demoninator.

So now I've just got n plus one to the nth power in the denominator.

And in the numerator, I've still got n to the n.

And the numerator I'm dividing by two to the n.

So I can put that in the denominator.

And in the denominator, I am diving by the two to

the n plus one so I can put that in the numerator.

I've got two to the n plus one divided by two to the n.

Everything except for a single factor of two cancels.

So what I'm left with here is, N to the N, over N plus one to the N,

times two. But this, I can combine to be the limit.

N goes to infinity of N over N plus one to the N.

And that's to get that times two.

But we already started that, the limit of this N over N

plus one to the N, as N approaches infinity, is one over E.

So, this whole

limit, is two over e and two over e is less than one.

So, by the ratio test, this series converges.

What if that two became a three?

Does the series n goes from one to infinity of n factorial over n divided

by three to the nth power. Converge or diverge?

Now, this series doesn't converge. Here's the argument that we used to

show that the sum of n factorial over n over two to the n converges.

Now we switched that two for a three.

We just figure out how this argument needs to be changed.

So let's replace this two here with a three.

And now the claim is that that series doesn't converge

anymore, but that it diverges.

Again, I should be applying the ratio test here, so I

am looking at the limit of the ratio of subsequent terms.

But this has some twos that I swapped out for threes.

Here I've got some twos that I need to swap out for threes.

And here I got some twos.

That I need to swap out for threes.

And there's some more twos that need to be replaced with threes.

And here we got three to the n plus one over three to the n.

So instead of multiplying by two, I'm now multiplying by three.

This two becomes a three.

And here's the worst part. This two becomes a three.

And three over e is not less than one. Three over e is bigger than one.

And because big L is bigger than one, the ratio test says that this series diverges.

Let me leave you, with a question.

Does the series, n goes from one to infinity of n factorial divided by n

over e to the nth power converge or diverge?

[SOUND]

[SOUND]

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