“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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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

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

Alternating series test.

[SOUND] Alternating series.

And it turns out that there's a particularly nice criterion for

when they converge.

So here's what normally goes by the name, the alternating series test.

So the setup is as follows.

I'm going to suppose that I've got a sequence,

the terms are decreasing and all of the terms are positive.

And the limit of the nth term as n goes to infinity is 0.

So these are the assumptions, and then in that case I may conclude that this

alternating series that I built from the sequence a sub n by multiplying

by -1 to the n+1 power, this alternating series converges.

Alternating series form a class of series for

which the limit test is practically the whole story.

So given these conditions, if the limit of a sub n is 0, then this series converges.

But we already know by just the limit test, that if the limit of the nth term

of this series remember is non 0, then the series diverges.

But to say that this limit is non 0 is exactly the same thing as saying that this

limit either doesn't exist or is non 0.

So alternating series are a situation where just determining the limit of

the nth term or at least the absolute value of the nth term,

tells the whole story.

If that limit is 0 then the series converges, and if that limit is non 0,

then the series diverges.

Now I already know that if the limit of the nth term is not 0,

the series diverges.

So, in this case, what do I have to show?

So this is a statement that I want to prove.

I start with some sequence which is decreasing, all the terms are positive.

The limit of that sequence is 0.

I want to conclude that this series, this alternating series converges.

Well, what does that even mean?

What does it mean to show that this series converges, right?

I just have to remember the definition of this is really something about the limit,

all right?

It's to say that the limit of this partial sum,

the sum k goes from 1 to n of -1 to the k + 1 times a sub k.

This sequence of partial sums should converge and

I'll usually denote that nth partial sum by just S sub n.

Well, let's look more carefully at these partial sums.

So let's look at the first few terms of the sequence of partial sums, right?

What happens when I just plug in a few values of n here, what do I get?

When I plug in n=1, it's not too interesting, right?

It's just the sum of the first term, which is just a sub 1.

When I plug in n = 2, that's a little bit more interesting, right?

That's the sum of the first two terms, which is a sub 1- a sub 2.

Why is the minus in there?

When I plug in k = 2 here, I get -1 to the third times a sub 2,

-1 to the third is -1.

So that's introducing a negative sign in front of the a2.

So this is a sub 1- a sub 2.

What's the third partial sum?

Well, the third partial sum is a sub 1- a sub 2 plus the next term in the series,

which is -1 to the fourth times a sub 3, it's plus a sub 3.

What's the fourth partial sum?

Well, that's a sub 1- a sub 2 + a sub 3- a sub 4,

I mean it's maybe you're not getting a lot of intuition here.

The fifth one is a sub 1- a sub 2 + a sub 3- a sub 4 + a sub 5.

It's a little bit hard maybe to tell what's going on.

So, instead of just looking at these numbers,

I could try to look at these partial sums on a number line.

And the key fact to remember here is that these a sub ns are decreasing.

So in something like the 5th partial sum, a sub 2 is smaller than a sub 1,

a sub 3 is smaller than both of these still, right?

That's going to help us to draw a nice picture of what these partial sums look

like on the number line.

So, the first partial sum, who knows where it is?

It's just a sub 1 and it's somewhere on the number line.

But then to get from that first partial sum to the next partial sum,

to get from S sub1 over to S sub 2, I just subtract a aub 2.

So here I am subtracting a sub 2, and

that lands me over here at the second partial sum,

because S sub 2 is a sub 1- a sub 2.

Well, how do I get to the third partial sum?

Where's the third partial sum?

Well, to get from S sub 2 to S sub 3, I just add a sub 3.

And the key fact here is that a sub n is decreasing, so

the amount that I'm moving to the right with a sub 3

is less than the amount that I was moving to the left with a sub 2.

So maybe here is S sub 3, the third partial sum.

And then to get from S sub 3 to S sub 4, I have to subtract a sub 4.

But again, the sequence a sub n is decreasing, so

a sub 4 is smaller that a sub 3.

So I'm moving less to the left than I was to the right to go from S

sub 2 to S sub 3.

So now I subtract a sub 4 and I land here at S4, the fourth partial sum.

And to get from S sub 4 to S sub 5, I'd be adding a sub 5.

But a sub 5 is even smaller than a sub 4, so maybe that's somewhere over here.

What do I notice?

Well, what I'm noticing here is that the even partial sums,

S sub 2 and S sub 4, are increasing.

And the odd partial sums, S sub1, S sub 3, S sub 5,

they're decreasing, could be a little more precise, right?

This sequence, the sequence S sub 2n-1 is decreasing.

This sequence, S sub 2n is increasing.

So these two sequences are monotone.

The S sub evens is increasing, but bounded above by S sub 1,

and the S sub odds are decreasing, but bounded below by say S sub 2.

So I've got monotone and bounded sequences.

So by the monotone convergence theorem, the sequence of even partial

sums is increasing and bounded above and therefore the limit exists.

And by the monotone convergence theorem,

the sequence of odd partial sums is decreasing but bounded below.

And consequently the limit of S sub odd exists.

Well, that's awesome or not?

What did I actually want to prove?

Well, I actually wanted that the limit of just the partial sums exists.

I don't really care about the sequence of even partial sums and

the sequence of odd partial sums.

I just want know about the sequence of partial sums, does it have a limit?

So how do I know that the odd and the even sequences converge to the same thing?

So what I know is that the limit of the even partial sums exist and

the limit of the odd partial sums exists.

But I just want to know about the limit of the partial sums all together, all right?

The even terms are getting close to something, the odd terms are getting

close to something, but are all the terms getting close to something?

Well, to analyze this, I'm going to look instead at this limit.

I'm going to look at the limit of the difference of the even and

odd partial sums, and there's two different ways to calculate this limit.

On the one hand, well, I can just calculate what this term is,

this is the sum of the first 2n terms, this is the sum of the first 2n-1 terms.

So if I add up the first 2n terms and then subtract all but the last term,

that's exactly the same then has just the 2nth term,

which in this case is -a sub 2n.

So this limit is the same as this limit, but what is this limit, right?

Well, I actually know what that limits equal to,

because I assume the limit of the nth term is 0.

So [LAUGH] my assumption, the limit of -a sub 2n is

the same as negative the limit of a sub n which is 0.

But now there's another way that I could calculate this limit.

This is the limit of a difference which is the difference of the limits provided

the limits exist.

And in this case, they do, right?

I'm assuming that these two limits exist.

So this is the limit of a difference, this is the difference of the limits,

but now, I just calculated this a moment ago to be equal to 0, right?

What I'm saying here is that the limit of this sequence and

the limit of this sequence differ by 0,

which is to say that this limit equals this limit, right?

Which is just to say that the limit of S sub n exists because the limit of

the even terms is equal to the limit of the odd terms.

So all of the terms together are getting close to something in particular.

And that's what I wanted, the limit of the partial sums exist and

that's exactly what it means to say that this series converges.

To say that a series converges is just to say

that the limit of the sequence of partial sums exists.

But wait, there's more.

Let's think back to this picture, right?

This diagram of the number line with the partial sums drawn on it.

Well the odd partial sums are decreasing, and their limit is some value L.

The even partial sums are increasing, and their limits the same thing, it's also L.

Because the difference between [LAUGH] even and odd partial sums are controlled

by just the terms in the sequence a sub k and those terms have limit 0.

So the even partial sums are increasing to L,

the odd partial sums are decreasing to L.

And what this means is that the value of this series is between these two, right?

Here's the value of the series.

Here's the series and all of the even partial sums are below it and

all of the odd partial sums are above this true value L.

This is one of the best reasons to care of our alternating series.

So not only is it really easy to determine if an alternating series converges, but

the whole story just boils down to whether the limit of a sub n is 0 or not.

But more than that,

I can now get explicit error bounds on the value of an alternating series.

If I compute the 2nth and the 2n-1th partial sum,

I know the true value of my series lands between these two values.

And this makes alternating series great for machines.

When you're calculating a partial sum, you're getting error bounds for free.

[SOUND]

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