“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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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

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 use the monotone convergence theorem.

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Well here is a complicated sequence.

It's a sequence that starts with its first term equal to one, and

subsequent terms, a sub n+1, we define it in terms of a sub n.

Do square root of a sub n+2.

This sequence is bounded above by 2.

Well let's see why.

Let's suppose that a sub n is between zero and two.

What do I know about the next term, a sub n+1?

Well then a sub n + 1 must also be non-negative, and

that's just because it's a square root of something.

But how big can a sub n + 1 possibly be?

Well, a sub n + 1 is the square root of a sub n, which could be as big as 2, + 2.

So a sub n + 1 can be as big As the square root of 2 plus 2, which = 2.

And this is great, right?

This is telling me that if I know that a particular term is stuck between zero and

two, then the next term is also stuck between zero and two.

And what do I know about the first term?

Well the first term is definitely between zero and two,

right, a sub 1 is between zero and two.

And that means that the next

term must also be between zero and two.

And that means that the next term, a sub 3, must be between zero and two.

And that means that the next term, which is a sub 4 must be between zero and two.

That means that the next term, this a sub 5 must be between zero and two and so on.

And what this actually is telling me is that no matter what n is,

a sub n is between zero and

two and the sequence is non-decreasing.

To show that the sequence is non-decreasing,

we're going to show that a sub n is less than or equal to the next term, a sub n+1.

Of course, I got a formula for the next term,

that's just the square root of a sub n+2.

So the question, is this true?

Is a sub n less than or equal to the next term, the square root of a sub n + 2?

And since a sub n is non-negative,

I can figure out when this happens just by squaring both sides.

So I'm wondering, is a sub n squared less than or equal to a sub n+2?

I'm going to subtract a sub n squared from both sides.

So I'm wondering when is negative a sub n squared plus a sub n+2 not negative.

Now I can factor this.

This is asking when is zero less than or equal to, how does this factor.

We gotta two there, so I'll write a 2, 1, 2 minus a sub n,

1 plus a sub n, is how this quadratic polynomial factors.

So the question is, when is this product non-negative?

Well in order for this product to be non-negative, either one of these is zero,

or they're both positive or they're both negative.

It can't happen that both these terms are negative.

The only possibility is that a sub n is 2 or a sub n is -1,

or both of these terms are positive.

And that happens when an is between -1 and 2.

So the upshot of this whole argument is that as long as a sub n is between minus

one and two, then a sub n is less than or equal to a sub n+1.

But I know that a sub n is between minus one and two because just a little while

ago we saw that a sub n was stuck between zero and two.

And consequently this sequence is non-decreasing.

What does that all mean?

So, got the sequence and the terms of the sequence are between zero and two and

that means that the sequence is bounded and the sequence is non-decreasing.

Means the sequence is monotone.

So we got a monotone bounded sequence.

And by the monotone convergence theorem,

that means that the limit of the sequence exist.

We can try to get some numerical evidence to, I guess the limit.

So the first term in the sequence which is defined to be one.

To get the next term, I use this recursive definition.

I'll add 2 and take the square root, and that gives me the a sub 2 term,

and that's the square root of 3, which is about 1.73.

Now to compute the next term, what do I do?

Well, using this formula I'll add 2 and then take the square root of all that.

That will give me the a sub 3 term.

And that's about 1.93.

Now to compute the a sub 4 term, right, it's the same sort of process.

I'll add 2 and then take the square root of all that.

That will give me the a sub 4 term.

And that's approximately 1.98.

And if I keep on going we might believe then

that the limit of a sub n as n approaches infinity is in fact two.

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