“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

数列

欢迎参加本课程！我是 Jim Fowler，非常高兴大家来参加我的课程。在这第一个模块中，我们将介绍第一个学习课题：数列。简单来说，数列是一串无穷尽的数字；由于数列是“永无止尽”的，因此仅列出几个项是远远不够的，我们通常给出一个规则或一个递归公式。关于数列，有许多有趣的问题。一个问题是我们的数列是否会特别接近某个数；这是数列极限背后的概念。

- Jim Fowler, PhDProfessor

Mathematics

Let's control the size of sequences.

[SOUND]

[MUSIC]

I want some terminology, some language, for us to be able to talk about

sequences that don't get too big or don't get too negative.

The word that we'll be using is bounded.

Let me give a precise definition.

So precisely, to say that a sequence is bounded above,

means that there is some real number M, that's the bound,

so that for any index, that term in the sequence is no bigger than M.

We can think about this graphically.

Well, here's the graph of some sequence.

Each of these dots represents a term in the sequence, and

the height of the dot represents the value of that particular term.

And, I positioned these terms in sort of a compressed way.

You'll notice that my labels here, on the n axis, aren't all equally spaced.

All right, I'm squishing them together, over here, so

I can fit the entire future of this sequence on a single sheet of paper.

Now the sequence is bounded, and what that really means is that

these terms never exceed some bounding value.

Here I've picked some value M.

No term of the sequence is above this horizontal line.

This horizontal line is representing an upward bound for this sequence.

I can also bound a sequence from below,

meaning that the sequence doesn't become too negative.

To say that a sequence is bounded below, means that there is some real number M,

that's the bound, so that for any index n,

the nth term in the sequence is no smaller than that bound M.

Sometimes, a sequence will be bounded both from above and from below.

Well, in that case, we just say that the sequence is bounded.

So a sequence is bounded exactly when it is bounded both above and below.

Now it's worth pointing out, that the definition for bounded below

involves a number M and the definition for bounded above includes a number M.

But, it's very unlikely that the upper bound and the lower bound are the same.

Of course, they could be the same, right,

if the sequence is just the constant sequence.

But in general, a bounded sequence is going to have a different upper bound and

lower bound.

So we've got our precise definitions.

Now it'll be fun to make up some sequences, and

try to figure out whether those sequences are bounded.

For example, let's think about this sequence.

The sequence a sub n = sin of n.

Is this sequence bounded above, bounded below,

bounded, neither bounded above nor below?

Well let's think about it, what do we know about sine?

Sin, no matter what I plug into sine,

it's not bigger than 1 and it's no smaller than -1.

Consequently, the sequence is bound above by 1 and bounded below by -1.

This is an upper bound, and this is a lower bound for that sequence.

And since the sequence is bound both above and below,

I can just write that the sequence Is bounded.

Now, you don't want to get the idea that this 1, and this -1,

are the only choices for upper and lower bonds.

I mean, I could also say that the sequence is bounded above by 17.

That would be accurate to say.

So there's lots of choices here, for this upper bound.

But in any case, in this example, the sequence is bounded.

The sequence sine event is bounded.

Let's do another example.

So let's look at a sequence, b sub n = n times

sin(pi times n, over 2).

Is this sequence bounded above, bounded below, bounded,

neither bounded above nor below?

How can we think about this?

Well this is sine of pi times a whole number over 2.

What are the possible values for this sine term?

Well this thing could be 0, it could also be 1.

It could also be -1.

And it'll be very large inputs, very large n, for which sin(pi times n, over 2) is 1.

And for those very large inputs, b sub n is just n times 1.

There'll be other very large inputs for which sin(pi times n, over 2) is -1.

And for those very large inputs, b sub n is going to be -n.

So that means that this sequence isn't bounded above and it isn't bounded below.

It's neither bounded above nor below.

Then, I can write out a more formal argument.

Let me try to write out a formal argument that this sequence is not bounded above.

But to make that argument precise,

I'm going to try to convince you that it's not bounded above by M.

But I'm not going to tell you what M is.

So I'm going to try to write down an argument that shows it

can't be bounded above by M, and since it can't be bounded above by an arbitrary M,

it's just not bounded above.

There's no upper bound.

So how do I know that this sequence isn't bonded above by M?

Well the trick is to pick some other number.

I'm going to pick some number that looks like this,

1 followed by a whole bunch of 0s, followed by a 1.

But I want to pick that number so

that it's bigger than your purported upper bound.

And let's call that number big N.

So no matter what M you pick, I can find a whole number like this,

1 followed by a bunch of 0s, followed by a 1, which is bigger than M.

Now what do I know about b sub big N?

Well this is N times sin(pi times big N, over 2).

But if you think about it a bit, sine of a number like this times pi over 2 is 1.

So for this particular value of big N,

b sub N is just N times 1, is just N.

But what do I know about big N?

Big N is bigger than M.

So that means that b sub N is bigger than M,

but that means that M can't,

Be, An upper Bound.

Big M can't be the big M that appears in the definition of bounded above.

And because this sequence isn't bounded above by an arbitrary big M,

it's just not bounded above.

There's no choice of M, for

which this sequence can be said to be bounded above by M.

That might seem to formal, let's to try to do some numeric calculations just to get

a sense of what's going on here.

Numerically, what's going on here?

Well b sub 1 is 1 times sin(pi over 2).

That's just 1.

b sub 2 is 2 times sin(2 times pi over 2), which is sin(pi), which is 0.

It's just 0.

b sub 3 is 3 times sin(pi times 3, over 2).

Well sin(pi times 3, over 2) = -1, so b sub 3 = -3.

Well how do the rest of the terms go?

Let me just start writing them out.

So 1 is the first term, 0 is the next term, -3 is the next term.

If I plug in 4 for n, I get 0.

If I plug in 5, I get 5.

If I plug in 6, I get 0.

If I plug in 7, I get -7.

If I plug in 8, I get 0.

If I plug in 9, I just get 9.

If I plug in 10, I get 0.

If I plug in 11, I get -11.

And it keeps on going.

Every other term is 0.

And the non 0 terms are flip flopping in sign, in S-I-G-N.

So 1, 5, 9, and the next term, 13, are all positive.

But -3, -7, -11, and so on.

These terms are negative.

So this sequence isn't bounded above and it isn't bounded below.

If I go far enough out in the sequence, I can find terms that are very positive.

And if I go far enough out in the sequence,

I can find terms that are very negative.

[SOUND]

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