“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

级数

在这第二个模块中，我们将介绍第二个主要学习课题：级数。直观地说，将数列的项按照它们的顺序依次加起来就会得到“级数”。一个主要示例是“几何级数”，如二分之一、四分之一、八分之一、十六分之一，以此类推的和。在本课程的剩余部分我们将重点学习级数，因此如果你在有些地方感到疑惑，将会有大量时间来弄清楚。另外我还要提醒你，这个课题可能会令人感到相当抽象。如果你曾经为此困惑，我保证下一个模块提供的实例会让你感到豁然开朗。

- Jim Fowler, PhDProfessor

Mathematics

We need general techniques for determining when a series converges.

We've already seen an example of what's usually called the comparison test.

The original series that we looked at was this series.

The sum as n goes from 0 to infinity sin squared N divided by 2 to the N and

that series converges but remember why.

Let's recall how we prove that this series converges.

And what we did was to compare to a G metric series.

So, zero is less than or equal to Sin squared and

over two to the N is less then or equal to one over to two to the N.

And these inequalities just follow from the fact that sin squared is trapped

between 0 and 1.

But what do I know about the sum of 1 over 2 to the n?

The sum 1 over 2 to the n, n goes from 0 to infinity, converges.

This is a geometric series, I can even evaluate it.

Now what does that tell me about the original series that I'm studying,

what about the sin squared n over 2 to the n?

Well, the sequence of partial sums for

this series are all bounded above by the value of this series.

And the sequence of partial sums is nondecreasing

because these terms are all non-negative.

So I've got a monotone sequence which is bounded above.

That means the sequence of partial sums converges.

And that's just what it means to say that this series converges.

That was a bit quick, so let's generalize this argument,

try to formulate it as a broadly useful technique.

So here's the usual setup.

I'm imagine that I've got two series, I'll use k as the index,

k goes from 0 to infinity, of a sub k.

And another series, k goes from 0 to infinity, of b sub k.

And I'm supposing that 0 is less than or equal to a sub k,

is less than or equal to b sub k, for all k.

I'm imagining that the sum of the b sub k's converges.

So I'm going to be assuming that this series converges and

then I'm wondering about this series.

Does that series converge or diverge.

So to study that question we're supposed to look at the sequence of partial sums.

So that's s sub n is the sum of the terms,

k goes from 0 to n of a sub k.

Now, the whole question is whether this sequence converges.

because the convergence of this sequence is exactly what it means to say that

this whole series converges.

What kind of sequence is that?

Well, here's what I know.

I know that a sub k, is less than or equal to b sub k.

And that means if I add up, a whole bunch of a sub k's, say the a sub

k's between k equals 0 and n, that's less than or equal to adding up the b sub k's.

K goes from 0 to n.

Now this thing here, is just the nth partial sum.

Now what do I know about this sum?

Well, the rest of the b sub k's, are all non negative.

So, if I add up even more b sub k's, I'm just making this even bigger.

So what I can conclude is this.

That s sub n is less than or equal

to the value of the series k goes from 0 to infinity Of b sub k.

In other words, what I can conclude is that s sub n is

bounded above.

And it's monotone.

So let's check that.

I'm going to show that if m is bigger than n,

then s sub m is at least as big as s sub n.

Well what's s sub m?

S sub m is the sum k goes from zero to m of a sub k and

I'm sure that that's bigger than or

equal to the sum k goes from zero to n of a sub k.

But n is bigger than m so I can rewrite this sum, I can split this sum up.

This is the sum k goes from zero to n.

Of a sub k, plus the terms between n plus 1 and m, so

k goes from n plus 1 to m, of a sub k.

And that's at least as big, I hope, as the sum k goes from 0 to n, of a sub k.

Now, why is this true?

Well I've got the sum k goes from zero to n of ace of k on both sides.

then I got this extra term over here.

What I need to know is that this extra term is at least zero.

But that's true because all the ace of k are non negative.

And if you add up a whole bunch of non negative numbers then the result

Is non-negative.

So, this is telling me that the sequence of partial sums is non-decreasing.

So the sequence of partial sums is non-decreasing and bounded above.

So by the monotone convergence theorem, it converges.

Let's summarize that.

So here's what we've proved, if we've got two sequences

which are related in this way, that a sub k is between zero and b sub k.

And the sum k goes from zero to infinity of b sub k

converges then the sum

k goes from zero to infinity of a sub k also converges.

We can also say something when the some of the A sub Ks diverges.

Let's suppose that zero is less than or equal to A sub K.

Less than or equal to B sub K.

And let's suppose that the sum of the A sub Ks.

K goes from zero to infinity, diverges.

And I want to know what can I say about the sum of the b sub ks.

Well, if this series diverges, that means that the sequence of partial sums,

the sum k goes from 0 to n of a sub k that can't converge, right?

Because to say this thing converges is to say the series converges.

So, if this series diverges, this sequence must diverge.

But that sequence is monotone, for the same reason as before, right?

I'm adding up non-negative terms, so as I add up more terms.

Well, at least it's non-decreasing.

So the sequence is a monotone sequence and yet, that sequence doesn't converge.

So that means that the sequence of partial sums must be unbounded.

because if it were bounded and it was monotone, then it would converge, but

it doesn't converge.

So I've got an unbounded sequence.

What about the sequence of partial sums for

the series involving the b sub k's, right?

So the sequence of partial sums there is the sum k goes from 0 to n of b sub k.

This is even bigger than the sequence of partial sums of the a sub k's.

This sequence Is likewise unbounded.

Well, if this sequence is unbounded, then that sequence doesn't converge.

And if that sequence doesn't converge,

that means that the sum of the bk's diverges.

And that was fast, so it helped to see the convergence and

divergence statements together.

Well here's our standing assumption that 0 is less than or

equal to ak is less than or equal to bk.

The first thing that we so

is the series of b sub case converges then the series of the a sub case converges.

In the other hand if the series of

a sub case diverges then the series of b sub case diverges.

It's super important to keep tracking the directions of this implications.

Here I've got the convergence of bk implying the convergence of ak.

And here I've the divergence of ak implying the divergence of bk.

And hopefully that makes sense.

This is saying that as long as you've got nonnegative terms,

if you're below a convergent series, you converge.

And this is saying, as long as you got non-negative terms,

if you're above a divergent series, you also diverge.

And let me end by saying that the comparison test is just a ton of

fun to use.

It's the definitely my favorite test.

[SOUND]

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