“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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“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

Monotinicity matters.

[MUSIC]

Do you remember the statement of the alternating series test?

So here is what the alternating series test tells us.

If I've got a sequence, the terms of which are decreasing and they're all positive,

and the limit of a sub n as n approaches infinity is 0.

Then the alternating series that I build using the sequence a sub n,

the sum n goes from 1 to infinity of -1 to the n+1, times a sub n.

So this is an alternating series.

This alternating series converges.

I've been assuming that the sequence a sub n was decreasing.

And I used that to get that the even and the odd partial sums were monotone.

But what happens if I get rid of that condition?

What happens if I get rid of the condition that a sub n is decreasing?

Well, let's break the alternating series test by removing this condition.

I'm going to remove the condition that the sequence a sub n be decreasing.

They call this the Broken Alternating Series Test.

I'm just going to assume that the a sub n's are are positive and

that their limit is 0.

And if I just make this two assumptions,

is it then the case that this alternating series necessarily converges?

So to see how broken this really is, the question is this.

Can you think of an alternating series where the terms are going to 0, but

the series diverges?

Well, here's an example, it's the sequence a sub n.

And the definition depends on whether n is odd or even.

So if n is odd, it's 1 over (n+1) divided by 2.

And if n is even, it's 1 over (n/2) squared.

Well, what does this look like?

I'm going to use this sequence to build an alternating series.

And what does this alternating series then looking like?

Well, I plug in n=1, and that's odd.

So it's 1 over 1+1 over 2, so it's 1 over 1.

I plug in n=2, I get 1 over 2 over 2 squared, but

that's going to come with a minus sign.

I plug in n=3, it'll be 3 is odd.

So it's 1 over 3+1 over 2, it's a half.

I mean, here is some of the terms, I've just started writing them down.

So it's 1 over 1, and then minus 1 over 1 squared, plus 1 over

2 minus 1 over 2 squared, plus 1 over 3 minus 1 over 3 squared, plus 1 over.

I mean, what this really amounts to

is sort of weaving together two different series.

All right, the positive terms are the harmonic series,

and the negative terms in this alternating series are the p-series where p equals 2.

And the limit of the nth term is 0.

More specifically, the limit of the odd terms here is just the limit of 1 over n,

and that is 0.

And the limit of the even terms here is just the limit of 1 over n squared, and

that's also 0.

So putting together these two facts, just the limit of a sub n is equal to 0.

But the sequence, it's not decreasing.

If I look at the terms, just the first few terms of this sequence a sub n,

I mean these are not decreasing.

I mean look, one-half and then one-fourth.

But then one-third, right?

It's going up here, down here, up here, down here, up here.

It's not a decreasing sequence.

And the alternating series diverges.

So I want to look at the sequence of partial sums s sub 2n,

the sum k goes from 1 to 2n, -1 to the k+1 times a sub k.

Now the whole thing about this series is that it's weaving together the harmonic

series and a p-series where p equals 2.

So this partial sum can be written like this.

The odd terms are the harmonic series, and

the even terms are giving me this p series with p equals two.

The odd terms are positive and the even terms are negative.

So this partial sum is this.

It's the sum k goes from 1 to n at 1 over k minus the sum k goes from 1 to

n at 1 over k squared.

Now the whole point here is that the harmonic series diverges, and

this a converging p-series.

So that means by taking n large enough, I can make this harmonic series or

this partial sum for the harmonic series as large as I'd like.

But no matter how big I take n, this convergent p-series

never exceeds the true value of this p-series, which is pi squared over 6.

So what happens then, when n is really large, right?

This term becomes very, very large, as large as I'd like.

This term never exceeds the value of the convergent p-series with p equals 2.

And that means that the limit of these even partial sums is infinite.

And consequently,

this series diverges because the limit of the partial sums doesn't converge.

Well, this is a particular philosophy by which you can study the whole world.

If you want to know that car engines are important, well,

you take the car engine out and the car doesn't go.

So it must have been important.

Well the same [LAUGH] sort of game is being played here with the alternating

series test.

I took out a part of the alternating series test.

I've removed the assumption that a sub n is decreasing.

And I saw that the alternating series test didn't work anymore.

So although I don't emphasize it all the time,

the fact that the a sub n sequence is a decreasing sequence

really is important when you're applying the alternating series test.

[SOUND]

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