“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。

Loading...

From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

45 ratings

The Ohio State University

45 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。

From the lesson

交错级数

在第四个模块中，我们讲解绝对和条件收敛、交错级数和交错级数审敛法，以及极限比较审敛法。简而言之，此模块分析含有一些负项和一些正项的级数的收敛性。截至目前为止，我们已经分析了含有非负项的级数；如果项非负，确定敛散性会更为简单，因此在本模块中，分析同时含有负项和正项的级数，肯定会带来一些新的难题。从某种意义上，此模块是“它是否收敛？”的终结。在最后两个模块中，我们将讲解幂级数和泰勒级数。这最后两个课题将让我们离开仅仅是敛散性的问题，因此如果你渴望新知识，请继续学习！

- Jim Fowler, PhDProfessor

Mathematics

Let's rearrange.

[MUSIC]

The alternating harmonic series is a great example of a conditionally

convergent series.

Well, here is the alternating harmonic series, it's the sum and

goes from one to infinity if negative one to the n plus one over n.

And what we know bout this is that it converges

by the alternating series test but it doesn't converge absolutely.

because if you look at some of the absolute values,

you're looking at the Harmonic series which diverges.

And because it converges but

not absolutely, we call it conditionally convergent.

And what about the negative terms?

What if I just add up the negative terms?

Well, here is the first dozen or so terms of the alternating harmonic series.

One over one, minus a half, plus a third, minus a fourth, plus a fifth,

minus a sixth, and so on.

What if I just add up these negative terms, just the terms With even index.

What do I get?

Well, in that case, what I'm looking at is the sum, and

goes from 1 to infinity of 1 over 2n.

These are the even index terms from the alternating harmonic series.

What do I get if I add up all of these numbers?

Well, that's one half of what I get when I add up one over n and

goes from one to infinity but, that's half of the harmonic series, right.

That means, that this diverges.

>> What if I just look at the positive terms?

>> Well, in that case, I'm trying to add up one over one plus a 3rd,

plus a 5th, plus a 7th, plus a 9th >> All right,

I'm trying to figure out just how big is this?

What's the sum n goes from 1 to infinity of the 1 over odd numbers, 1 over 2n-1?

Does this converge or diverge?

Well the trouble is that 1 over 2n-1 is even bigger than 1 over 2n.

>> one over one is bigger than a half.

A third is bigger than a fourth,

a fifth is bigger than a sixth a seventh is bigger than an eighth.

So if this series diverges, than this series diverges as well.

The sum of just the positive terms in the alternating harmonic series diverges and

it somehow, that series has a finite value so in the alternating

harmonic series the negative terms diverge, the positive terms diverge.

They both diverge.

So what I've got is really two piles of numbers and if I take enough

from either pile, I can make a number that's as large as I would like.

This presents me with the following quick, strange opportunity.

Here's my goal.

I'm going to rearrange the terms of the alternating harmonic series

to get a new series, same terms, just different order.

But now my new series, when I evaluate it, will have value, 17.

I'll keep picking up positive terms until I exceed 17.

Ok well, here we go.

Here's a number line.

Here's zero.

Here's my goal, 17.

We're trying to pick up numbers from this pile so

that I can move all the way past 17.

So I could just get started, right.

I pick up one, and that gets me a little bit closer to 17.

I'll pick up the next number, that's a third, that gets me a bit closer to 17.

I'll pick up a fifth That gets me a little bit closer to 17.

I'll pick up a seventh.

And that gets me even a little bit closer to 17.

I mean the trouble, of course, is that these numbers are getting smaller, but

I know that this series diverges.

So if I keep taking numbers from this pile,

I can move as far to the right as I like And indeed, it happens that the sum end

goes from one to ten to the 14th of one over two N minus one.

Is bit bigger than 17.

It's 17.1.

So, this is how I'm going to start.

I'm trying to write down the same terms as the alternating harmonic series but

I want a series now that converges to 17.

And this is how I'll start.

I'll write 1/1 + a third + a fifth, all the way to 1 over 2 times 10

to the 14th- 1, and that'll land me just to the right of 17.

And now, we'll use some of the negative terms.

The sum of these terms from this positive pile Was just about 17.1.

So, if I pick up a half from the negative pile, and

I subtract a half now, that moves me over to about 16.6.

Now we'll take some of the positive terms again.

Of course I've already used up a lot of positive terms, but

there's definitely more there that I can add.

Because this series diverges, I've only taken away a finite piece of it.

So there's more yet to grab.

It's just the numbers are really big.

Or really small, rather.

But in any case, there's more terms from this positive pile to add.

And if I add enough of them, I'll eventually move past 17 again.

Maybe I'll end up at say 17.001 or thereabouts.

And some more of the negative ones.

I'll take away this quarter.

I'll subtract a quarter from here, and now my 17.001,

or thereabouts maybe moves over to a little bit less than 17, say 16.751.

And I'll just keep on doing this.

I can add more positive terms again and move myself back to the other side of 17

and I'm never going to run out, right, because this pile of numbers is infinite.

I mean, this series diverges.

So I'm going to keep moving back and forth past 17.

And in the limit I'll get 17.

So does that mean that this sum is equal to 17?

No, no.

In fact we're going to see that this series is equal to the natural log of two.

What we're seeing here is the first glimpse of a theorem.

It's a Rearrangement Theorem, here's how it goes.

Suppose that (L) is some real number, you get to pick out, and

you got a conditionally convergent series.

In this case, I'm calling it the sum angles from one to infinity of A sub n.

So, (L) is the real number that you picked, and

you're given this conditionally convergent series.

Then that sequence A sub N can be rearranged to form a new sequence,

B sub N.

So the B sub N sequence contains all the terms of the A sub N sequence,

just in a different order.

Well that rearranged sequence B sub N.

If you form a series out of it the sum N goes from one to infinity of B sub N.

The value of that series is L.

And you picked L.

What this is saying?

It's saying that if you're given a conditionally convergent series, you could

rearrange the terms so that that theorem sums to any number that you would like.

In light of this theorem, we have to be careful about how we think about theories.

Order matters.

A series is a list of numbers to sum, In a given order.

It's not just a pile of numbers that you add up.

The numbers are coming at you in a given order.

[SOUND]

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