“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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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

The comparison test again.

[MUSIC]

Well here's something that happens.

Well maybe you're analyzing a couple series and

the first thing that you do is try to apply the limit test and

you find that in both cases the limit of the nth term is 0.

So at least these series aren't diverging for

an obvious reason like the limit of the nth term is non zero.

So you've got two series that may or may not converge.

But even though the limit of a sub n is 0 and the limit of b sub n is 0,

it could be that the limit as n goes to infinity of a sub n over b sub

n might be some number L which is positive.

What does that really mean?

Well, one way to think about that is that it's sort of saying something like this.

It's saying that a sub n, b sub n, are almost multiples of each other

like a sub n is almost a multiple of b sub n at least when n is really big.

I can be more precise with epsilons.

So let's set epsilon equal to L in this case.

Epsilon is going to be a positive number, but

I'm assuming that my limit L is positive.

So let's set epsilon equal to L.

Then the definition of limit says what?

It says that there's some big N so that whenever little n is greater than or

equal to big N, the distance between the thing I'm taking the limit of:

a sub n over b sub n, and my limit L, is less than epsilon.

And in this case, epsilon is L.

Now, that lets me compare a sub n, and b sub n.

Well, how so?

Let me make this assumption that the a sub ns and the b sub ns are non-negative.

I'm going to want that because I want to put the comparison test in a moment.

Then I can simplify this a bit.

Instead of making this claim, I can just get rid of the absolute value bar and

it's still true, right.

a sub n over b sub n minus L is less than L,

you can add L to both sides of that inequality.

And I get this that a sub n over b sub n is less than 2 L.

And now I can multiply both sides by b sub n and that's okay because

b sub n is not negative so it doesn't change the direction of this inequality.

And that tells me that this is for

the large values of little n, a sub n is less than 2 L b sub n.

Now all these pieces are really setting up a comparison test.

How's that going to work?

I got to remember this is only true for the large N but that's OK.

Well suppose that I knew that this series,

the sum of little n goes from 1 to infinity of b sub n converged.

Well then I would know that this series,

the sum little n goes from 1 to infinity if 2L b sub n also converges, right?

I can multiply convergence series just by some number.

But now I'm in a position to apply the comparison test.

Granted this statement is only true for large values of little n but

that's okay, right, because convergence only depends on the tail.

So this statement then implies that this series,

the sum of the a sub n's, converges, because this is bigger than a sub n.

I mean, at least for large values of little n.

So I'm getting a theorem that's telling me if I've got two series

of non-negative terms, and this series converges, then this series converges.

Provided that this is true.

Let me summarise that.

So, if I've got a sub n around that negative,

b sub n around that negative, the sum of the b sub ns converges.

And this limits statement that the limit of the ratios between the a sub n and

the b sub n is some finite value L, which is positive.

Then, I can conclude that this series,

the sum of the a sub n then goes from 1 to infinity converges as well.

Now let me exchange the roles of a and b.

So I'd like to be able to start with the assumption that the series of the a sub ns

converges and then conclude that the series that b sub ns converges.

But actually that's the exact same statement, right?

Watch this.

If I just replace this limit with this limit, now I'm in the exact same situation

except now b sub n's and a sub n's roles are exchanged.

Right? And that means that if I know that this

series converges, then I know that this series converges as well.

Now, I'll put it all together.

Well, since the convergence of a sub n implies the convergence of b sub n,

and the convergence of b sub n implies the convergence of

a sub n in the presence of this limit statement,

equivalently this limit statement, I can simplify this a bit.

I can say that if I've got 2 sequences of numbers, a sub n and b sub n,

both non negative and this limit exists and is equal some number bigger than zero.

Then this series converges

if and only if this series converges.

The sum of the b sub ns converges if and only if the sum of the a sub ns converges.

This conversions test has a name.

This is called the Limit Comparison Test, and it's one of the situations

when two series, in this case the sum of the b sub n's and

the sum of the a sub n's share the same fate.

They either both converge or they both diverge.

[SOUND] [BLANK AUDIO]

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