“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

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 not start with the zeroth term.

How do I sum this geometric series?

It's the sum, k goes from m to infinity of r to the k.

So this starts out with a term r to the m and

then the next term is r to the m + 1.

And it keeps on going like that.

It's a geometric series but the first term is r to the m not r to the zero.

I remember how to sum the series that starts, not with m, but with 0.

Well in that case, to evaluate the sum k goes from 0 to infinity of r to the k.

We already studied that and that is one over 1- r

provided the absolute value of r is less than one.

I want to somehow change that zero back into an m.

There's a general result that I can evoke.

Some constant times the series, which adds up all of the a sub

k's is equal to the series that adds up c times the a sub of k's.

This is of course assuming that these series make sense.

I'm assuming that this series converges in order to assert this equality.

So in this case I'll multiply by r to the m.

So let me write that down.

I'm going to multiply both sides by r to the m.

For the r to the m times the sum, k goes from zero to infinity

of r to the k is r to the m over 1- r.

Now, I can use these result.

Remember, that this result is assuming that these series converges.

So I'm always going to be working with the assumption that the absolute value of

r is less than one in order to guarantee that this series converges.

Let's apply the result.

So, if I apply the result in this case, what do I get?

Well, here r to the m is playing the role of c and r to the k is a sub k.

So, this can be replaced by this, which mean this is equal to what?

Is equal to the sum k goes from zero to infinity,

of r to the m times r to the k.

Which I could rewrite as the sum k goes

to 0 to infinity of r to the m + k.

The balance on the series can be rewritten.

What do I mean by that?

Well, let me write out what this series looks like.

I'm just going to write up the first few terms of the series.

The k equal to 0 term is r to the m,

the k equals one term is r to the m + 1.

The k equals two term is r to the m + 2 and so on.

So I could rewrite this series as the series that I'm interested in.

This series ends up just being the sum

k goes from m to infinity of r to the k.

And if I go all the way back, right, I figured out that that series

has this value, of course assuming that the absolute value of r is less than one.

But all told them I can conclude that this series is

equal to r to the m over 1 minus r as long as the absolute value of r is less than 1.

So that's a useful formula.

But even more useful than the formula is the method that we use for driving it.

We used this fact to derive that formula.

But this fact, while it looks like the Distributive Law,

is more than just the Distributive Law.

What I mean is that the Distributive Law says this.

It says that C (a0 + a1 +...+ an)

is = c times a sub zero + c times a sub one + c

times a sub 1 +...+ c times a sub n.

It is something about finite sums.

I could even write it using summation notation, right?

The Distributive Law says that c times

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

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

That's the Distributive Law is telling you.

But contrast that we this statement.

This isn't just a statement about finite sums.

This is a statement about infinite series.

So how do I justify something like that?

Well, it's more than just the Distributive Law.

Let me show you.

So I could write down c times the limit as n

approaches infinity of the sum k goes from 0 to n of a sub k.

And by definition, this is c times the infinite series

k goes from 0 to infinity of a sub k.

But I know something about limits.

A constant multiple of a limit is the limit of that constant multiple times

whatever I'm taking the limit of.

So this thing here is equal to the limit

as n approaches infinity of c times

the sum k goes from 0 to n of a sub k.

And now, I can apply the distributive law because this is just a finite sum.

It's a finite sum where n is maybe very big, but it's still just a finite sum.

So this is the limit, n goes to infinity of the sum

k goes from zero to n of c times a sub k.

So, this equality is the Distributive Law.

Now, I've got the limit of this sum and

that is the infinite series k goes from 0 to infinity of c times a sub k.

So, let's look at what happened here, right?

This is just the definition of the infinite series.

The fact that these are equal is a property about limits.

The fact that these are equal is the Distributive Law.

And the fact that these are equal is the definition of infinite series.

So the fact that a constant multiple times

an infinite series is the infinite series of that constant multiple of the terms

is more than just the Distributive Law at play.

It's really the Distributive Law plus a fact about limits.

[SOUND]

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