“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。

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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。

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

Absolute convergence.

[SOUND] What do I even mean by absolute convergence?

Well here's the definition.

All right, so definition, the series.

I'll write down a series.

The sum n goes from one to infinity of a sub n.

Converges absolutely.

So this is the term that I'm defining, defining absolute convergence.

I'm going to say the series converges absolutely if what happens.

If the series, just the sum

N goes from one to infinity of the absolute values of the a sub ns.

Just plain old converges in the usual sense.

What's an example of this?

A while ago, let's look at this series.

The sum n goes from one to infinity of the absolute value of sin n over two to the n.

That series converges, and you could prove that by doing a comparison test with

the geometric series one over two to the n.

All right, so that series converges.

And that means that this series,

the sum n goes to one to infinity of just sin n over two to the n.

No more absolute value now, just sin n over two to the n,

that series converges absolutely.

Or we might say is absolutely convergent, just because this series,

the sum of the absolute value of this quantity, converges.

But with a name like absolute convergence, you'd hope that an absolutely convergent

series is also just convergent in the regular sense of converging.

So, that's a theorem.

If the sum of the absolute values of the a sub ns converges.

Then just the sum of the a sub ns converges.

In other words, if a series converges absolutely.

Then it just plain old converges.

Let's prove that.

Okay, yeah.

So how's this proof going to go?

Well the proof is going to start out with my assuming

that the series converges absolutely.

Meaning I'm going to assume that the sum of the absolute values of the a sub ns

converges.

And then somehow, I want to do something.

I don't know what yet.

So I've got some cloud with question marks.

Then I eventually want to conclude that the original series converges.

So what goes in these question marks, right?

How do I get from this assumption to this conclusion?

Well, the first thing I can note is that if this series converges.

Then the sum of twice those terms also converges.

Why is that important?

Well, here is the key fact.

This is sort of a trick if you like, but what it's saying is that zero is

less than or equal to a sub n plus the absolute value of a sub n is less than or

equal to twice the value absolute value of a sub n.

Why is this even true?

Well, let's think about the possibilities, maybe a sub n is negative, and

if a sub n's negative.

Then I've got a negative number plus well,

the absolute value of that negative number.

These two things cancel and I just get zero, and then certainly this is true.

Zero is less than or equal to zero is less than or

equal to twice some positive number.

What if a sub n were positive?

Well then I'd have zero is less than or equal to a positive number plus itself.

because the absolute value of a positive number is just that same number

is less than or equal to twice that same number.

Well, in that case these are equal.

Right, I've got a positive number plus itself that's the same as twice that

positive number.

So if a sub n's positive this is true.

If a sub n's negative this is true.

If a sub n's zero, this is just zero less than or

equal to zero plus zero less than or equal to two times zero.

So whether an is positive, whether it's zero.

Whether it's negative.

No matter what happens, this is a true statement.

Now we can apply the comparison test.

So now by the comparison test, this series converges.

Right? The sum of these converges.

The sum angles from one to infinity of a sub n plus the absolute value of a sub n.

Well that's term wise less than this convergent series, and

all the terms are non-negative.

So that means this series converges.

But now, our original series can be written in terms of this series and

the sum of the absolute values at the a sub ns.

So what I've got is this series converges this series converges.

This series converges. But the difference of these two series,

is the series that I'm originally interested in right.

And the difference of convergent series converges.

So that means can't the series just the sum of the a sub n

converges which is exactly what I wanted to show.

So we proved the theorem.

If a series converges absolutely then it just plain old converges and

that really justifies the terminology..

I'm calling this, well everybody calls this absolute convergence.

Because it's a strong kind of convergence.

Absolute convergence implies convergence.

So absolute convergence is even stronger than

just regular old convergence.

[SOUND]

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