“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。

Loading...

From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

46 ratings

“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

N log n.

[SOUND]

[MUSIC]

The harmonic series diverges.

But the sum of the reciprocals of the squares converges.

P series with p equals 2 converges, right?

So this series converges.

The harmonic series diverges.

So this series diverges.

But what's something in between, right?

Where's the boundary between convergence and divergence?

Well, here's an example of a series that sits in between.

It's the sum n goes from 4 to infinity of 1 over n log n.

Now, 1 over n log n is less than 1 over n

because n times log n is bigger than n, right?

A bigger denominator makes the fraction smaller.

And because n squared is bigger than n log n,

1 over n squared is less than 1 over n log n.

So this thing is bigger than something that converges,

smaller than something that diverges.

So it's really a question, right?

Does this series converge or diverge?

We can use Cauchy condensation.

So let's analyze this series using Cauchy condensation.

So I'll write down the condensed series.

It's the sum n goes from 2 to infinity of 2to the n times the 2to the nth term.

So the 2 to the nth term means i put it 2 to the n times log 2 to the n in

the denominator.

Okay, so that's the condensed series.

This is the original series.

Cauchy condensation tells me that these two series share the same fate.

They either both converge or both diverge.

So it's enough to figure out what's going on with this series.

And look.

I've got a 2 to the n in the numerator and a 2 to the n in the denominator, so

I can cancel those.

And I can rewrite this as the sum, n goes from 2 to infinity.

Just 1 over log 2 to the n.

Now, I can use properties of logs to simplify this.

This is the sum, n goes from 2 to infinity of 1 over n times log 2.

That's just because log 2 to the n is n times log 2.

And then, we'll look at this.

This is 1 over log 2 times the sum,

n goes from 2 to infinity of 1 over n.

But this is bad or really great news, depending on your attitude.

This thing here is the tail of a harmonic series.

So that means that the condensed series diverges.

And because the condensed series diverges, this original series must also diverge.

If you like integrals, you can use an integral test.

Okay, so let's do this with the integral test.

So I could look at this integral,

the integral from 4 to infinity of 1 over x times log x dx.

That's a suitable function to consider.

By definition, this integral from 4 to infinity just means the limit as n

approached infinity of the integral from 4 to big N of 1 over x times log x dx.

But now, how do I evaluate that definite interval?

Well, I happen to know this.

The derivative of log of log of x is what?

By the chain rule, this is the derivative log, which is 1 over, evaluate at

the inside, log of x, times the derivative of the inside function, which is 1 over x.

Well, here, I've got 1 over x times log x.

Here, I've got some other function.

Here, I've got an anti-derivative for this integrand.

So that's enough for me to be able to calculate this definite integral

using the fundamental theorem of calculus.

So this is the limit as n approaches infinity of, what is this?

This is telling me an anti-derivative.

So log log x evaluated at 4 and N.

And then I could plug in N and plug in 4 and take the difference.

This is the limit N goes to infinity of log log N minus log log 4.

Now what is this limit?

Well, it's growing very slowly, but it is in fact running off to infinity, right?

By choosing N big enough, I can make log N as large as I like, and

that means I can also make log log N as large as I like.

So this limit is in fact infinity.

That mean that this integral diverges.

That means that the original series diverges as well.

So that diverges, but maybe if we mess around with it a bit,

we can make it converge.

So our original question was whether or not this series converged or diverged.

And now we've seen, both by using condensation and

the integral test that this series diverges.

But remember back to our original story, right?

We were thinking of this series that's somehow sitting in between

these two series.

But now, where's the boundary between convergence and divergence, right?

This series converges, these two series diverge.

So let me try to fit another series in between these two series.

Here's an example.

It's the sum, n goes from 4 to infinity of 1 over n times log n, and

that log n is squared.

Does that series converge or diverge?

We can use condensation.

So let's write down the condensed series.

And in this case this is the sum,

n goes from 2 to infinity of 2 to the n times the 2 to the nth term.

So 2 to the n times log 2 to the n squared.

How do I evaluate this series, right?

If I can determine the convergence of this series,

I've determined the convergence of the original series.

Good news, this 2 to the n and this 2 to the n cancel.

So now I'm left with the sum, n goes from 2

to infinity of 1 over log 2 to the n squared.

What else can I do there?

I can use properties of logs again.

So this is the sum, n goes from 2 to

infinity of 1 over n times log 2 squared.

I could factor out the 1 over log 2 squared.

So this is 1 over log 2 squared times the sum,

n goes from 2 to infinity of 1 over n squared.

This is really great, right?

Because what do I know?

This is a p series with p equals 2.

This series converges.

Well, that means that this condensed series converges.

That means the original series converges.

So we've got a convergent series and a divergent series.

It's worth comparing these two series.

This series,

which we were asking about, we now know converges by Cauchy condensation.

And we've already seen a little while ago that this series diverges.

So again, we could try to play the same game,

which is something that we try to fit in between these two.

Well here's a question.

Does this series, sort of fits in between these two series,

does this series converge or diverge?

This is the series n goes from 4 to infinity of 1 over n

times log n times log log n.

Well, I'll leave it to you to analyze this series.

[SOUND]

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.