“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

How large is large enough?

[SOUND] The definition is limit says

that to get within epsilon of L,

I just have to go past the big Nth term.

To guarantee that you're within epsilon, how big does N need to be?

We can actually compute this in some cases.

Consider the sequence a sub n equals n +1/ n+2.

So what's the limit?

The limit of the sequence has an approaches infinity is 1.

Let me draw a picture of this.

So here's a number line.

Let's put 0 all the way over here.

Let's put 1 right here.

Now I've got a bunch of terms of the sequence, right?

So here's the first term, here's the second term.

Here's the third term and so on.

And as I go out further and further in this sequence,

the terms get closer and closer to 1.

And the question is how far do I have to go out in the sequence

to guarantee that I'm within some epsilon of 1?

Now let's suppose that I want to be within a hundredth of 1.

How big does N have to be?

So what I want to do is find a value for N so that whenever n is bigger than or

equal to N, I get that the nth term of my sequence is

within one-hundredth of my limit 1, right?

But this is telling me that the nth term is within epsilon,

epsilon being one-hundredth in this case, of my limiting value, 1.

But I can rewrite this.

Instead of writing it this way, I could instead write that a sub n should be

between ninety nine-one hundredths and a hundred and one-hundredths.

And to be between 99 and hundred and

one-hundredths is exactly the same thing as being within a hundredth of 1.

Now I got a formula for a sub n.

So I could instead write this is ninety nine-hundredths, the formula for

a sub n is n+1 over n+2 is less than a hundred and one-hundredths.

So what I'm trying to do is figure out how big I need N to be so

that whenever n is bigger than N, I know that both of these inequalities hold,

meaning that my nth term is really within a hundredth of 1.

Well one of these inequalities comes for free.

This inequality here comes for free,

because n+1/n+2 is always less than 1, right?

The numerator here is smaller than the denominator.

So this thing being less than 1,

in particular this thing is less than hundred and one-hundredths.

So I get this inequality for free.

This inequality however requires a little bit of work.

I could solve here by multiplying both sides by n plus 2 and

by 100 and I end up finding that n needs to be at least 98.

So as long as I choose a value for big N which is bigger than 98,

that guarantees that this inequality holds.

This inequality holds automatically,

that tells me that my inth term really in within hundredth of 1.

This is pretty, it's awesome it's pretty cool that we can tell if your

past the 98th term in this sequence.

Then you're within a hundredth of 1.

And there's nothing special about a hundredth.

If you want it to be within a billionth of 1 you just have to go much further out in

the sequence and then you get there right?

And no matter how close you want to be to 1 if you go far enough out in the sequence

you'll be that close.

And that's exactly what it means to say that the limit of this sequence is 1.

[SOUND]

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