“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

级数

在这第二个模块中，我们将介绍第二个主要学习课题：级数。直观地说，将数列的项按照它们的顺序依次加起来就会得到“级数”。一个主要示例是“几何级数”，如二分之一、四分之一、八分之一、十六分之一，以此类推的和。在本课程的剩余部分我们将重点学习级数，因此如果你在有些地方感到疑惑，将会有大量时间来弄清楚。另外我还要提醒你，这个课题可能会令人感到相当抽象。如果你曾经为此困惑，我保证下一个模块提供的实例会让你感到豁然开朗。

- Jim Fowler, PhDProfessor

Mathematics

Cauchy Condensation

[MUSIC]

Let's think back to the harmonic series.

Well, here's the harmonic series.

And we proved that it diverges.

And how did we do that?

Well, we did this grouping trick.

Here's the beginning of the harmonic series, and it keeps on going.

And what we saw was that we've got a one here and a half here.

This third and this fourth can be grouped together.

And the next four terms can be grouped together.

And then the next eight terms will be grouped together, and so on.

But this group here is at least a half,

because it's two terms that are each at least as big as a fourth.

This next group here is also at least a half,

these are four terms that are at least an eighth.

And then the next group of 8 terms is at least a half, and

the next group of 16 terms is at least a half.

So this is even worse than adding up 1 plus a half plus a half plus a half plus

a half plus a half.

And that series diverges, so we are able to show that the harmonic series

diverges by doing this grouping into piles with sizes of powers of 2.

This isn't just some one-off trick, this is a technique we can generalize and

then apply broadly.

And we used it for the harmonic series to show that the series diverged.

We can also use the same kind of thinking to show that a series converges.

The name for this trick is Cauchy Condensation.

Here's how it goes, at least when using it to prove convergence.

So let's suppose that I've got a decreasing sequence and

that all of the terms are positive.

I want to know, does this series,

the sum K goes from one to infinity of a sub K, converge or diverge?

I can group them by powers of two.

So what I mean is I'll put a sub one by itself, I'll group together a sub 2 and

a sub 3.

I'll group together a sub 4, a sub 5, a sub 6 and a sub 7.

I'll group a sub 8 alongside a sub 9, a sub 10,

a sub 11, a sub 12, a sub 13, a sub 14, a sub 15.

And then I'll group a sub 16, a sub 17.

I'm going to keep on going until I get to a sub 31.

And then I have to start a new group at a sub 32.

Now I'll overestimate the groups.

Well, how big is a sub 2 plus a sub 3?

The key fact here is that the sequence of the a sub k's is decreasing.

That means that a sub 3 is smaller than a sub 2.

So if I add a sub 2 and a sub 3, the result is less than twice a sub 2.

What about this next group?

What's a sub 4 plus a sub 5 plus a sub 6 plus a sub 7?

Well a sub 5, a sub 6 and a sub 7 are all less than a sub 4,

again because the sequence is decreasing.

So if I add these four numbers up,

the result Is less than if I just multiply the fourth term by four.

What about the next eight terms?

Well these terms from a sub 9 through a sub 15 are all smaller than a sub 8.

So this eight terms all together are less

than just eight copies of a sub 8.

And it keeps on going, right?

The next 16 terms, if I were to group them all together,

would be less than 16 copies of a sub 16, and so on.

Let me write down a precise statement of what we're doing, so

here's how I can write down the grouping in symbols.

So we'll get the sum of the a sub ks,

k goes from 1 all the way up to 1 less than a power of 2.

So, 2 to the n-1, say.

So, this will be like adding up the first 7 terms, or

the first 15 terms, or the first 31 terms.

It's a collection of terms where I can actually do the grouping.

Well, this is less than or

equal to than the sum of what happens after I do the grouping, right?

Which is 2 to the k times a sub 2 to the k.

This is what I get when I do this grouping.

k goes from 0 all the way up to n- 1.

Why is this significant?

So we were originally interested in studying this series and

determining whether this series converged or diverged.

And now we're led to studying this series.

The sum k goes from 0 to infinity of 2 to k

times the 2 to the kth term in the sequence.

I'm going to call this series the condensed series associated to

this series.

Now what if the condensed series converges?

So if this condensed series converges, right, then what happens?

What we're calling, I know, right,

this condensed series is overestimating the original series.

So if I'm trying to study something about the partial sums of the original series,

what I know now is that those partial sums for the original series are bounded above

by the value of the convergent condensed series.

I also know that the partial sums for the original series are increasing.

And that's just because all of the a sub ks are positive.

So as I add up more and more of the a sub ks, I'm getting a bigger and

bigger number.

So the sequence of partial sums is increasing.

That means a sequence of partial sums is bounded above and increasing and

therefore convergent.

And what that means then is that the original series that I'm interested in

converges.

Let's summarize the theorem.

So here's a statement of the Cauchy Condensation test.

If you've got a sequence which is decreasing and all the terms are positive,

then the series sum k goes to one to infinity a sub k

converges if the condensed series converges.

That's what we just proved.

And in fact, it's an if and only if.

So the fate of the condensed series is exactly the same as the fate

of the original series.

[SOUND]

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