“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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From the course by The Ohio State University

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

审敛法

在第三个模块中，我们学习用于确定级数是否收敛的各种审敛法：特别地，我们将说明比值审敛法、根值审敛法和积分审敛法。

- Jim Fowler, PhDProfessor

Mathematics

The ratio test.

[MUSIC]

The ratio test, asks us to look at the limit of the ratio of subsequent terms.

Here's a precise statement.

To the question is whether the sum n goes from 1 to infinity of a sub n converges or

diverges.

And we're going to want to assume that all of the terms that I'm adding up.

All of the a sub n's are positive in order that I can use the comparison tests.

I'll compute the limit of the ratio of subsequent terms.

So I'm setting big L to be the limit as n approaches infinity of a sub

m plus 1 over a sub n.

And then there's three possibilities.

Maybe that limit, maybe big L is less than 1.

And in that case, the ratio test says that the series converges.

Maybe that limit is bigger than 1,

in which case the ratio test says the skew diverges.

And maybe that limit is equal to 1,

in which case the limit test is simply silent.

It might converge, it might diverge we just don't know

the ratio test doesn't tell us any information.

When L is equal to 1.

Why does this work, what if the limit is less than 1?

So I'm trying to figure out what happens when big L is less than 1.

I will show the series in that case converges of big L is less than 1.

I can pick some very tiny value of upsilon so

that even big L plus upsilon is less than 1.

That I can use Epsilon and the definition of limit.

so there must be a big n so whenever little n is at least big n.

The thing I getting the limit of a sub n plus one over a sub n

must be with Epsilon of L.

So in particular a sub n plus one over a sub n must be less than L plus Epsilon.

Now, how is this fact helpful?

Well, I'm trying to calculate this series.

I'm trying to sum a sub n, n goes from 1 to infinity.

But I can break it up into two pieces when n is smaller than big N and

when n is bigger than big N.

Well, here's the piece when n is smaller than big N.

It's just a finite sum.

I'm not going to have any trouble adding those numbers up.

But then I've got the rest of the series that I have to add up.

I have to add up a sub N + a sub N +1 and so on.

And this is the hard part.

All right? Because I've got dot, dot, dot forever.

But what I know now is that I can control the ratio between subsequent terms,

maybe not here, but at least here.

I know that a sub big N plus one is less than R times a sub big N.

And I know that a sub big N plus two is less than R times

a sub big N plus one which itself is less than R times a sub big N.

So all of that is to say that this thing, here, is,

well not equal to this, but this is an overestimate of what was there before.

So now I've got that this series is, at least,

less than this first part plus the rest.

But what do I know about the rest?

Well, the rest of this thing is now a geometric series.

And I know that this geometric series converges because little r is less than 1.

So, I can replace that with this convergent geometric series.

And in light of this, all right this is really now setting up a comparison test.

I'm really saying that the sequence of partial sums is bounded by this.

And since the sequence of partial sums is monotone because all

the terms that I'm adding up are positive I then know that this series converges.

And what if the limit is bigger than one?

So I'm trying to figure out what happens when big L is at least 1.

And I want to show that in that case the series diverges.

Well, if big L is bigger than 1, then I can choose some tiny epsilon, so

that big L minus epsilon is also bigger than 1.

Then I can use that epsilon in the definition of the limit.

So there's some big N, so that whenever little n is bigger than or equal to big N,

this ratio, the thing I'm taking the limit of, is within epsilon of the limit L.

So in particular, this ratio is bigger than big L- epsilon.

Why is that significant?

What I'm trying to calculate is this series,

or I'm at least trying to figure out if it converges or diverges.

And I could start just adding up terms like a sub 1 + a sub 2,

eventually I get to a sub N-1, then a sub N + a sub N+1 and so on.

I keep up adding up terms, making it limited partial sums strictly speaking.

But I know that all of these terms are positive.

So if I throw away a bunch of terms, well this isn't equal to this anymore, That

this is now larger than this thing where I've thrown away a whole bunch of terms.

I'm going to start with the a sub big N term.

What else do I know?

Because a sub little n plus 1 over a sub n is bigger than this

whenever little n is bigger than equal to big N, I can underestimate this.

All right, this is saying that a sub big N + 1 is larger than r times a sub big N.

Telling me that this is bigger than r times this,

which is also bigger than r times this.

Which means this is larger than r squared times this, this term here,

a sub big N + 3 is larger than r cubed times this.

What I'm really saying is that this infinite series

is at least this infinite series.

But what kind of infinite series is this?

Well, this is just a geometric series, right?

That's the geometric series the sum k goes from 0 to infinity of

r to the k times a sub-big N What do I know about that?

If that diverges.

I can make the partial sums for

this as large as I'd like, as long as I go far enough out in a sequence.

Well, that means that this series also diverges.

And what is the ratio to say if a limit is equal to 1?

When is L = 1 the series might converge or

might diverge the ratio of tests is just silent in that case.

It doesn't tell us any information.

And why not?

When L < 1, then I could pick a value with Epsilon so

small that even L + Epsilon < 1.

And when L was bigger than 1, I could pick a value of epsilon so

small that even big L minus epsilon was bigger than 1.

But when L is equal to 1 there is no hope for me being able to choose a small

enough epsilon so that I can control the ratio between subsequent terms.

To eventually all be less than 1 or eventually all be bigger than 1.

So I'm not going to succeed in comparing my series to a geometric series.

So that's the ratio test but finally, a warning.

The ratio test tells us to compute this,

the limit of the ratio between subsequent terms.

It's important to emphasize that that limit is not calculating this.

What the ratio test is telling us is not the value of this series.

What the ratio test is providing information about

is whether this series converges or diverges.

That's what this big L is helping us find.

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