“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

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

Let's sum a series.

[MUSIC]

I want to evaluate this series carefully, so the claim here is that

the sum K goes from 0 to infinity of 1 over 2 to the K is equal to 2.

But what does that even mean?

Well what it really means is this.

It means that the sum k goes from zero to n of one over two to the k

is close to two, whenever n is big.

This sum makes sense because this is just a finite sum.

It's just telling me to add up one over two to the k for

the values of k between zero and n.

So this at least makes sense.

What is close to two mean?

Well it means it's close to be to two.

And how big does n have to be.

We'll it has to be true whenever n is larger than sum fixed

big value that depends on how close you want the thing to get to two.

That is the same thing but in terms of limits.

So more precisely, I'll define the sequence S sub n,

sequence of partial sums to be the sum.

K goes from zero to N of one over two to the K.

And the claim that this series has value two,

just means that this sequence has limit two.

It means that the limit as n approaches infinity of the nth partial sum Is 2.

How do I evaluate that limit?

Well, let's look for a pattern.

So, S sub 0, well that's the sum K goes from 0 to 0 of 1 over 2 to the k.

So that's just 1 over 2 to the 0 which is just 1.

S sub 1, well that's the sum K goes from 0 to 1, of 1 over the 2 to the K.

That's 1 + 1 over 2 the first power which is a half.

S sub 2, well that's the sum K goes from 0 to 2 of 1 over 2 to the K.

That's 1 + 1/2 + 1 over 2 squared which is 1/4.

Now I could simplify that a bit, I could write that as 1 + 3/4.

S sub 3, that's the sum, k goes from 0 to 3 of 1 over 2 to the k.

So that's 1 + 1/2 + 1/4 + 1 over 2 to the 3 which is 1/8.

And I could add 1/2, 1/4 and 1/8 and get 7/8.

And then I could compute S sub 4, right, and

that'll be 1 + 1/2 + 1/4 + 1/8 + 1/16 and that'll be 1 + 15/16.

And then maybe you'd be beginning to the beginnings of the pattern here, right?

I've got a half, three-quarters, seven-eights, fifteen-sixteenths here.

So it's perhaps believable, I mean this isn't approved,

there's just a little bit of evidence.

But the pattern it's suggesting itself that the nth partial sum is 1 +,

it has to be a number just a little bit less than 1 and

it's the corresponding power of 2 in the denominator and 1 less in the numerator.

So I could write that as 1 + 2 to the n minus 1 over 2 to the n.

Then, I could simplify that a little bit.

I could split that up as 1 + 2 to the n over 2 to the n minus 1 over 2 to the n.

And I could write that as 2 minus 1 over 2 to the n.

I haven't proved that, but hopefully that formula seems believable.

In any case, armed with that formula, we can evaluate the limit.

Here we go.

I want to to take the limit of s sub n as n approaches infinity.

And I got the formula for the nth partial sum.

So that's the limit as n approach infinity of 2- 1 over 2 to the n.

That's a limit of the difference which the difference the limit is part of

the limit exist.

So this is the limit just of 2 and as n approaches infinity

minus the limit of 1 over 2 to the n as n approaches infinity.

This limit of a constant which is is that constant minus this limit of a quotient

but look at what happens here, the denominator is very very large.

I can make the denominator as large as I like, right?

2 to the n can be made very very positive.

One over a very large number is very close to zero.

So this limit is in fact 0.

And that means that the limit of the partial sums is just two.

That means that the series converges to 2.

And I can not only see that algebraically by using that limit, but

I can also see it geometrically.

Geometrically, I can draw a picture like this where I've got 1/2,

1/4 of the square, 1/8 of the square, 1/16 of the square,

1/32 of the square, 1/64 of the square.

And all these pieces fit together to build one unit square.

And what that's showing me is that the sum k goes from 1 to

infinity of 1 over 2 to the k.

Well, that really is one.

So the sum k goes from 0 to infinity of 1 over 2 to the k, well,

that's the first term plus all the rest of these terms.

So that's 1 + 1.

That must be 2.

[SOUND]

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