“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

Let's look at ratios.

[MUSIC]

We've already seen how to determine the convergence of a geometric series.

For example, does the series the sum n goes from 1 to infinity

Of 1 over 4 to the n, converge or diverge?

Well, this series converges, right?

By which test, by the geometric series test,

if you like, because the common ratio, right,

this common ratio between the terms is one-fourth And that's less than 1.

What if we modify the series somewhat?

So what if instead of looking at this series, we looked at the series,

the sum n goes from 1 to infinity, not 1 over 4 to the n,

but n to the 5th over 4 to the n?

That series is not a geometric series anymore, but

it ends up at least approximately looking like one.

Let me just give some names to these things.

Let's say that a sub n is n to the fifth over 4 to the n.

And what I'm trying to determine is the sum,

n goes from 1 to infinity of a sub n, right?

I'm probably not going to actually determine its value, right?

But I'm just trying to anser the question.

Of whether or not that series converges or diverges.

Now, if I plug in really big values for n,

I can get a sense of what these terms look like.

All right, a sub 1000 is well,

hard to write down, but I can at least write down 1000 to the fifth

over 4 to the 1,000th power and it's hard maybe to see what that number looks like.

a sub 1,001, like a 1,000 in 1st term is

1,001 to the 5th over 4 to the 1,001st power.

And what's the ratio between these terms?

Right, how big are these terms compared to each other?

Now if this were actually a geometric series the ratio between subsequent terms

would be the same.

And this isn't a geometric series, but let's see if we're close.

So a sub 1001 divided by a sub 1000,

well a sub 1001 is 1001 to the fifth power over four to the 1001st power.

Divided by a 1000, which is 1000 to the 5th divided by 4 to the 1000th.

I could rewrite this,

because I've got a fraction with fractions in the numerator and the denominator.

I could rewrite this as 1001 to the 5th, times 4 to the 1000th power,

divided by A 1000 to the fifth times and the denominator and

numerator write that in the denominator 4 to the 1000 and fifth power.

Now I should split this up and think about this two pieces separately.

Alright how big are these two pieces well a 1001 of the fifth power

divide by 1000 to the fifth power this piece here is pretty close to one.

And this piece here is, well it's exactly one fourth.

So if I've got a number that's close to one, and

I'm multiplying it by a number that's close to a fourth, this fraction is,

it's not equal too, but it's about a fourth.

And if you think about it, there's nothing special about 1,000 here,

except that it's really big.

So the ratio between subsequent terms, at least if those terms are far

enough out in the sequence, that ratio is close to 1/4.

Can I make this precise?

I'll use a limit.

What I'm saying is that the limit as n approaches infinity

of A sub n + 1/A sub n In this case is equal to a quarter.

That if I choose n big enough, I can make that ratio close.

How close?

As close as you want to a quarter.

I can be very explicit in this case.

I could pick an explicit value of epsilon, let's just pick 0.1 It turns out

that whenever little N is bigger than 15, ace of N, plus one,

over ace of N is in fact within epsilon of my limiting value of a quarter.

In particular, if Ace of N plus one over Ace of N is within point one of a quarter,

For that also means that a sub n,

plus 1 over a sub n is less than a quarter plus 0.1.

And that's enough to set up a comparison with the geometric series.

I could multiply this inequality by the positive number a sub n.

And I find out that a sub n,

plus 1 is less than a quarter Plus .1 times a's of n.

Well, this also means, that a's of n plus 2 is less than a quarter

plus .1 times a's of n plus 1.

But, a's of n plus 1 is less than a quarter plus .1 of a's of n.

So, a's of n plus 2 is less than a quarter + 0.1 squared times a sub n.

Now this fact, with n replaced by n + 2, tells me that a sub

n + 3 is less than (1/4 + 0.1), 0.1 times a sub n + 2.

But a sub n + 2 is < (1/4 + 0.1) squared times a sub n.

So a sub n + 3 is < (1/4 + 0.1) cubed times a sub n.

Well this works in general, right?

What I'm really showing here Is what?

I'm showing that a sub n+k is less than

a quarter +0.1 to the kth power, times a sub n.

And this is true whenever n is at least 15.

So in particular, I'm showing that a sub 15+k

Is less than 1/4 + 0.1 to the kth power times a sub 15.

The key point here,

is that 1/4 plus epsilon is still less than 1 Well the series

that we are interested in is the sum angles from 1 to infinity of a sub n.

And that series converges precisely when the series

end goes from 15 to infinity of a sub n converges.

The first fourteen terms here,

that aren't included there, don't affect convergence at all, alright?

That's just adding fourteen more numbers to this But how do we analyse this?

We'll use the deal, right,

I got bound on the size of these a Sabines as long as n we use fifteen, i got this.

So the sum n goes from fifteen to infinity of a Sabin

is equal to the sum of k goes from zero to infinity Of a sub 15 + k, right?

The zero term here is just a sub 15 which is exactly the same as the first

term here, k = 1 is a sub 16 here and that's exactly the next term here.

So these are really the same series but how do I know about these series?

Well, that series converges y because the sum k

goes from 0 to infinity of a quarter plus .1 to the k times a's of 15 converges,

why does this series is convergence affect these convergence?

Well, this series as terms are bigger than the series terms and

this series terms are all positive.

So this follows from the comparison test that if this series converges,

then this seires converges.

But now why does this series converge?

Well this series converges by, say, the geometric series test.

And how does the geometric series test affect this?

Well this is a geometric series this a sub 15 is just a constant and

the common ratio is a quarter + 0.1.

And a quarter + 0.1 < 1.

So the common ratio,

being less than 1, means that this geometric series converges.

Means that this series converges.

Means that the original series that we're studying converges.

This is a specific case of a very general procedure.

I'm studying the convergence of some series.

The sum, n goes from 1 to infinity of A sub n.

I want to know if that converges or diverges.

Let's suppose that all the terms in the sequence, A sub n, are positive.

What I'm suggesting that we should try to do is to look at the limit

as n approaches infinity of a sub n + 1 / a sub n, all right?

This series is almost surely not a geometric series, right?

Most series aren't a geometric series.

But I could try to see is it close to a geometric series.

What is the ratio between subsequent terms?

Is it close to anything eventually?

Or is that what the limit is asking me?

So let's suppose that this limit does exist.

Let's suppose that this limit equals L, and let's suppose that L is less than 1.

Well then what I'm suggesting is that we pick some small epsilon How small?

Well I want epsilon so small that L plus epsilon is less than one.

If L is less than one I can definitely pick epsilon small enough that even if

I add epsilon to L I'm still less than one.

Now, I'm going to use epsilon in this limit, alright?

The fact that this limit equals L means that there's some big N.

So say find big N, so that whenever little n is bigger than or equal to big N,

a sub n plus 1 over a sub n should be within epsilon of L and

in particular this should be less then L plus epsilon.

And why does that help?

Well this inequality implies that a sub N plus k

is less than L plus epsilon, to k, times a sub N.

Where does this come from?

Well it really does come from just applying this a ton of times.

What this is saying is that as long as n is at least N,

then the ratio between subsequent terms is less than this factor.

And that means that I can overestimate the N+K term

by this factor of the K power times just whatever the N term is.

Now how do I use this?

Well, the sum k goes from zero to infinity of this,

of l plus epsilon to the k times a sub n converges.

And why is that?

Well, this is a geometric series with common ratio l plus epsilon.

that common ratio is less than one.

That's exactly why I picked epsilon so small up here.

So I've got a geometric series with common ratio less than one, so

this series converges.

But then by the comparison test, I know something about this series.

I know that the series n = N to infinity Of a sub n also converges.

Because these terms are all overestimated by these terms.

And consequently the original series, the sum n goes from one to

infinity of a sub n, also converges,

because I can get this from this just by adding on [INAUDIBLE] extra terms.

We can summarize this.

So, here's a statement of what's normally called the ratio test.

And it goes like this.

I want to know whether the series Engel's from one to infinity of a sub n

converges or diverges.

And I'm only going to study series that terms of which are all positive and

I need that because my argument was using the comparison test and

the comparison test requires series with at least nonnegative terms.

All right, then I'm going to look at the limit

n goes to infinity of an+1/an, the limit of the ratio of subsequent terms.

Let's suppose that limit exists and equals L.

Then there's three possibilities.

If that L < 1, Then, this series converges.

And it's converging because I can do a comparison test, between this series and

a convergent geometric series.

If that limit is bigger than 1, then this series diverges.

And I know that because again if this limit is bigger than one,

I can compare maybe not the whole series, but at least after awhile for some,

pass some big nth term I can compare this series to a divergent geometric series.

Then finally maybe L is equal to one and

in that case the ratio test is silent.

It doesn't tell us

any information either way.

[SOUND]

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