“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。 注意：此课程的注册将在2018年3月30日结束。如果您在该日期之前注册，您将可以在2018年9月之前访问该课程。

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From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

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“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。 注意：此课程的注册将在2018年3月30日结束。如果您在该日期之前注册，您将可以在2018年9月之前访问该课程。

From the lesson

审敛法

在第三个模块中，我们学习用于确定级数是否收敛的各种审敛法：特别地，我们将说明比值审敛法、根值审敛法和积分审敛法。

- Jim Fowler, PhDProfessor

Mathematics

The ratio test is awesome.

[MUSIC]

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.

All right, a of 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 the infinity, what's a sub n plus one is

going to replace this n with n + 1 that's n+1 factorial divided..

By (n+1) to the n+1th 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

the 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.

I've got an (n+1) factorial in the numerator and

an n factorial in the denominator.

So this n factorial cancels everything except for the n + 1 term here.

So this is the limit as n approaches infinity

of just n + 1 times n to the n divided by,

n factorial is gone now n plus one to the power n plus one.

Well, now I've got n + 1 in the numerator, and

the power of n plus one in the denominator.

So, I can use this to change this n plus one reduced to nth power.

So this is the limit.

As N approaches infinity of N to the N over N+1 to the N.

If you like, I can rewrite this a bit too.

I could write this as the limit as N approaches infinity.

And instead of separately raising the numerator and

denominator to the Nth power, could write this as n over n plus one to be nth power.

But how do i evaluate the limit?

Well, analyze this limit if you really love l'Hopital's Rule

you could just apply l'Hopital's Rule.

I don't really like l'Hopital's Rule that much so

instead i'm just going to recall a useful fact.

In fact this might have 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.

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.

Lets set an = n!/((n/2) to the nth).

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

and 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 + 1 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 + 1 And

plus 1 factorial over and plus 1 over 2 to the n plus 1 power

divided by a's of n which is n factorial over and over 2 with the n power.

And it's kind of a mess, because I've got fractions in the numerator and

denominator.

So I can go and 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 n factorial.

Most of those terms cancel except the n plus one.

So I can rewrite that as just n plus one in the numerator.

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

I can write this as n to the n divided by two to the n.

So this is n to the n divided by two to the n, and the denominator here, or

the n factorial goes away.

But I can rewrite this as n plus one to the n plus one power

divided by two to the n plus one.

Now I can keep simplifying this.

I've got a 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 denominator.

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

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 the denominator, I'm dividing by two to the n + 1.

So I can put that in the numerator.

Now I've got two to the n + 1 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 /(n+1) to the n times two.

But this, I can combine to be the limit n goes to infinity

of (n/n+1) to the n and I still got that time two.

But we already saw that the limit of just (n/n+1) 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 and

divided by three to the n 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 should 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'm looking at the limit of the ratio of subsequent terms.

But this has some two's that I swapped out for three's.

Here I've got some two's that I need to swap out for three's,

and here I've got some two's that I need to swap out for three's.

Here's some more two's that need to be replaced with three's, and

here we got three to the n + 1 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 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]

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