“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

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

Absolute convergence.

[SOUND] As we've seen practically all the convergence tests that we have

at our disposal have made the assumption that the terms in the series are positive.

Or at least non-negative.

I want to prove that the sum N goes from 1 to infinity of

minus 1 to the Nth power over N squared converges.

So, how am I going to do this?

I can't apply the usual convergence tests

because not all the terms are non-negative.

Look, here I've written out some of the terms.

Its minus 1 plus a 4th, minus a 9th plus a 16th, minus a 25th I can only

apply the comparison test if the terms were non-negative.

And that's not the case here.

But I can think about absolute convergence.

I can use the theorem that absolute convergence implies just

regular old convergence.

So, if I can just prove that this series converges absolutely,

then I know that it converges in just the usual sense.

So, let's try that.

What I know is this.

The sum n goes from one to infinity of the absolute value of minus

1 to the n over n squared.

Well that's exactly the same thing as the sum n goes from 1 to infinity

of just 1 / n squared and that's a p series.

With p = 2, and because 2 > 1,

this p series converges.

Now what does that mean in the original series I care about?

That means that the sum n goes from 1 to infinity of (-1) to the n /

n squared converges absolutely, because the sum of the absolute values converges,

and consequently by the theorem, it just plain old converges.

That's often how this is going to work.

Suppose you want to analyze this series.

The sum n goes from 1 to infinity of a sub n, you've been given this task.

Well the first thing I'd suggest you do is the limit test.

Take a look at the limit of an as n approaches infinity.

because if that's not 0, then you know your series diverges, you're done.

But let's suppose the series passes that test, then what do you do?

Well then you can hope that the terms in the series,

the a sub n's are all greater than or equal to zero.

because if you've got a series, all of whose terms are non-negative,

then you can apply all the usual conversions tests.

But if that's not the case, if you're in a situation where some of these terms

are positive, some of these terms are negative, what do you do?

Well, then I'd recommend that you apply all of our old conversions tests.

Not to this series directly, but to this series.

The sum n goes from 1 to infinity of the absolute value of the a sub n's.

What I'm suggesting that you do is try to prove that this series converges

absolutely.

Because if you know that this series converges absolutely,

then you know that this series just plain old converges.

So absolute convergence is an important idea not because every single

convergent series converges absolutely.

That's not even true there are series that converge but don't converge absolutely.

Nevertheless a lot of series do converge absolutely.

So an easy way to prove convergence is to prove absolute convergence and then to

use the theorem that absolute convergence just implies regular old convergence.

And it's going to be successful a lot of the time.

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