“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。

Loading...

From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

46 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。 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

Coarseness.

[NOISE]

[MUSIC]

Often, we've been asking for things on the nose.

Like in this formulation of the comparison test.

For all n, 0 less than or equal to a sub n less than or

equal to b sub n, if this series, the sum of the b sub ns,

converges, then this series, the sum of ace of n's converges.

When you formulate it this way, I'm asking for this inner quality for

hold on the nose that ace of n be between 0 and b of n for all n.

But now, we're beginning to see that we don't need to be quite so rigid.

Well, one thing we've seen,

is that convergence depends only on the tail of the series.

So something like this.

This assumption here.

For all n zero is less than or equal to a sub n is less than b sub n.

This can be weakened.

I can get away with just assuming this.

That for all large values of little n a sub n is between zero and b sub n.

And I can replace the statement that a sub n is less than b

sub n with something like the limit of the ratio.

Still there may be assumption that the terms are not negative, but

instead of this inequality, I can rewrite that

to just say that a Sabin over b Sabin this ratio is between zero and one.

And the point am thinking about this ratio or I can just get rid of this and

instead of thinking about this ratio being between zero and

one I can just ask that the limit of this ratio be some finite value, L.

And at that point,

[LAUGH] I've written down one direction of the limit comparison test.

So, we've replaced the comparison test,

which is really a pretty rigid statement with something much more flexible.

So yeah, the limit comparison test

is a coarser version of the just the regular old comparison test.

But by coarser, I actually am claiming as a sort of philosophical claim that

the limit comparison test is actually better than the comparison test.

And it's better for two reasons.

Well first of all,

the limit comparison test is better than just a comparison test in practice.

Beyond any sort of theoretical advantage I mean,

the limit comparison test just ends up being easier to apply.

It's easier to verify this condition than it is to cook up some sequence

of b sub ns which satisfy this, and the sum of the b sub ns converges, say, right?

This is much harder to satisfy than just some limit statement like this.

Well, let's look at an explicit example in practice where this really is much easier.

So here's an example of a series.

Let's analyze the conversion for this.

The sum n goes from 1 to infinity out of this fraction

n2 + the square root of n- n / n4 + n2- n.

Does this series converge or diverge?

What I can do is build a sequence out of this.

So I'm going to call the terms here a sub n, and then I'm going to look at this

other sequence, sequence b sub n equals 1 over n squared.

And I'm going to compute the limit of the ratio between the terms in

these sequences.

So what's the limit as n goes to infinity of a sub n over b sub n?

We're dividing by 1 over n, is the same as just multiplying by the reciprocal,

so I'm really asking about this limit.

But this isn't so painful now, right?

because now, I'm multiplying n squared times this, so

this limit Is this fractions limit of this fractions?

n to the 4th plus lower order terms n squared times n squared

plus lower order things and the denominator is the same.

It's just n to the 4th plus lower order things.

In this case, n squared minus n.

So what's this limit?

Well, if I've got n to the 4th plus lower order terms divided by n to the 4th plus

lower order terms, this limit ends up being 1 so

the limit comparison test applies.

What I know is that this series,

the sum of the b sub n's converges because it's just the p series with p equals 2,

and that means as a result of the limit comparison test,

I can then conclude that this very complicated series also converges.

But it's not just that it's easier to apply.

All right, there's a second reason why this kind of course thinking

is really beneficial.

And the reason is that thinking this way actually gets us closer to

understanding what convergence really means.

All right, what's conversions all about?

Well, its certainly not about the first million terms in your series.

Conversions only depends on the tail.

And what conversions really depends on is how quickly the terms

are decaying to zero.

So something like the limit comparison test that

asks you to look at the limit of the ratio of the terms in the series

is really asking you to think about how quickly those terms are decaying to zero.

And that's good, right?

That means the limit comparison test isn't just a way to get the answers more easily.

The limit comparison test is really showing us something about what

convergence really means.

And that's the goal of mathematics.

The goal of mathematics isn't just answers, it's understanding.

Well, think back to that example again, right.

This example wasn't just answering a question.

This example is demonstrating this philosophy.

That I'm doing more than just answering the question of this

series of convergence.

I'm saying that there's a reason,

an understandable reason why this series should convert.

This series converges because it resembles

in the sense of this limit between the terms and series.

But this series resembles 1/n squared.

And this p series, where p equals 2 converges.

So by the limit comparison test, this series does, too.

But that's not just a calculation.

That's not just an answer.

That's a reason why something that looks like this ought to converge.

[SOUND]

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.