“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

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 is looking like a great test.

Does it always work?

[MUSIC]

Let's recall what the ratio test says.

I'm going to be considering a series, the sum of a sub n, n goes from 0 to infinity.

And I'm going to assume that all of the terms that I'm adding up,

all of the a sub n's are non negative.

So a sub n is greater than or equal to 0.

And the ratio test tells me to consider

the limit of the ratio of subsequent terms.

So I'm taking the limit, as n approaches infinity, of a sub n + 1 over a sub n.

And I'm calling that limit L.

And then what does the ratio test tell me?

Well the ratio test tells me that if that limit is less than 1,

then the series converges.

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

But if the limit is equal to 1, then the ratio test is inconclusive.

So what happens when L equals one?

Well supposedly, the ratio test doesn't say anything in that case, but

if we listen really carefully,

is there something that we can get out of the ratio test even when L equals one?

No!

Whoa, okay, but why not?

Well there are cases where L=1 and the series converges.

But then there are also cases where L=1 and the series diverges.

Can we actually see some of these examples?

Well, here's an example.

Maybe kind of a silly example, of where L=1 but

the series will end up diverging.

So let's try this.

Here is the example.

It's the sum n goes from 0 to infinity of the number 1.

[LAUGH] I mean, this definitely diverges, right?

If you keep adding up 1+1+1, right, that's not a finite number.

So, this is a divergent series.

But of course in this case, what's a sub n?

Well, a sub n doesn't depend on n.

It's just always 1.

And so the limit of the n plus 1th term over the nth term as n goes

to infinity is just 1 because a sub n + 1 and a sub n are both 1.

What about an example where the series converges even though L equals 1?

So I want an example where the L in the ratio test is 1 but

the series ends up converging, even though the ratio test doesn't tell us that.

So, let's see an example.

An example of this is the sum n goes from 1 to infinity of 1 over n squared.

This series, we've already seen to converge.

It converges by, let's say, the p-series test.

It's a p series where p equals 2.

But what is L in this case?

Well, the nth term in the series a sub n is 1 over n squared.

So L is the limit of a sub n + 1 over a sub n as n goes to infinity.

And in that case, that's the limit as n goes to infinity

of 1 over (n + 1) squared divided by 1 over n squared.

And that limit is indeed 1.

But how do I know that this limit is 1?

Well, to compute this limit, I could first rewrite this as the limit

as n approaches infinity of 1 over what?

I'm just going to move the n plus 1 squared, the denominator, so

it'll be n plus 1 squared over n squared.

Now I could expand out the n plus 1 squared.

This'll be the limit as n approaches infinity of 1 over n squared + 2n +1.

That's when I got when I expand n+1 squared, divided by n squared.

And now this is a sum here in the numerator so

I could split this up into three separate fractions.

This is the limit as n goes to infinity

of 1 over n squared over n squared + 2n

over n squared + 1 over n squared.

And I can simplify this a bit more too.

This is the limit as n goes to infinity of 1 over

n squared over n squared is 1, 2n over n squared is 2 over n,

and 1 over n squared, I'll just write as 1 over n squared.

Well what's this limit?

When n is very large, this is very close to zero and this is very close to zero.

So when n is very large this is 1 over 1 + a number close to 0 plus a number

close to 0.

This limit is 1, which is exactly what I'm claiming.

So when we say that the ratio test is silent when L equals 1.

It's not for a lack of understanding.

There really are a series that converge and a series that diverge and

in both cases L equals 1.

[BLANK

AUDIO]

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