“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

Why positive?

[MUSIC]

Most of our convergence tests so far have been assuming that the terms in

the series are positive or at least non-negative.

Yeah, when we were thinking about the comparison test, the ratio test,

even the root test.

All right, in all of these convergence tests we're trying to determine whether

some series converges.

But we are making this assumption that all of the terms are at least non-negative.

Why is this?

Let's think about why the ratio test, in particular, requires this condition that

the terms be non-negative when you're analyzing this series.

Remember what the ratio test tells us to do.

It tells us to look at this limit and

say if this limit is less than one, then the series converges.

And it shows that by doing a comparison with this geometric series.

So indeed, I mean, if you're given the ratio test you look inside,

what you see is that the proof basically amounts to doing

a comparison test with a geometric series.

And the comparison test is really what requires this condition,

that the terms to be non-negative.

Okay, so most of these convergence tests are at some level

just reducing the problem down to a comparison test.

But that just raised another question.

Why does the comparison test require non-negativity of the terms?

Well, think back to how we proved the comparison test.

It's going to become clear that this condition,

the non-negativity of the a sub ns, is extraordinarily important.

Remember what we did to prove the comparison test.

If you open up the comparison test,

it amounts to the monotone convergence theorem.

It's really just an application of the monotone convergence theorem.

Let's remember how we did this, right?

How do we prove the comparison test?

So, one direction went like this.

I'm imagining I've got a series, the sum of the b sub ns converges,

and the a sub ns are trapped between b sub n and 0.

Then I want to conclude with the sum of the a sub ns converges.

Well, because the a sub ns are all non-negative,

that tells me that the sequence of partial sums is non-decreasing, all right?

This is where I'm using the crucial fact that the a subs ns are not negative

is to get that the sequence of partial sums is non-decreasing.

And because b sub n is greater than or equal to a sub n, I also know that

the sequence of partial sums is bounded by the value of this convergent series.

And consequently because the sequence of partial sums is monotone and

bounded, that tells me that the limit of the sequence of partial sums exists.

In other words, the sum of the a sub ns converges.

So non-negativity was important for

the convergence tests because they relied on the comparison test.

And non-negativity was important for the comparison test

because the comparison test is applying the monotone convergence theorem.

And I need non-negativity of the terms in the series

in order to know that the sequence of partial sums is monotone.

Now, that raises a question.

How can I analyze a series if the terms in the series aren't all non-negative?

And indeed, that very question is going to be a major theme for what is to come.

What are we supposed to do with series when some of the terms are positive and

some of the terms are negative?

[SOUND]

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