“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 add up the terms in a geometric progression in general.

[MUSIC]

Pick a value for r, here's the series I want to evaluate.

>> The sum, k goes from zero to infinity of sum number r raised to the kth power.

These series have a name.

>> These things are called geometric series.

We've already seen this when r equals one half.

>> Right.

So, if I plug in r = 1/2, I look at this geometric series,

this series we've already evaluated, it converges and its value is 2.

>> Or certain other values of r,

we can also figure out what's happening with this series right away.

>> Instead of considering one half, let's think back to this original geometric

series and let's plug in 1 for r.

If I plug in 1 for r, does this geometric series converge or diverge?

Well, it's not too hard to think about.

Let's start writing down what this series looks like.

The series would be the 0 of term, it would be 1 to the 0 which is 1,

that k=1 term would be 1 to the 1 which is 1.

The K=2 term is equal 1 squared which is 1 and

at this series is just 1+1+1+1 and so on.

Does this series converge or diverge?

This series diverges because the limit of the partial sums is infinity.

>> Something bad also happens when r equals negative one.

>> So yeah, instead of looking at r equals 1, let's plug in r equals -1 and

see what sort of terrible thing happens.

So if r equals -1,

then I can start writing down the first few terms of this series as well.

When I plug in k = 0, I get (-1) to the 0 power, which is 1.

When I plug in k = 1, I get (-1)

to the first power, which is -1.

Then I plug in k = 2, and I get (-1) squared, which is +1.

Then I plug in k = 3, and I get -1 cubed, which is -1.

So what this series look like is 1-1+1-1+1-1+1-1 and so on.

Does this series converge or diverge?

More precisely, what's the limit of the partial sums.

But when I add just the first term, I get 1.

When I add the first two terms I get 0.

When I add the first three terms together I get 1.

When I add the first four terms together, I get 0 and this pattern continues.

When I add the first five terms together I get 1.

When I add the first six terms together, I get 0.

The partial sums are flip flopping between 1 and 0, and

does a sequence one, zero, one, zero, one, zero does that sequence converge?

No.

And because the sequence of partial sums doesn't converge,

the original series doesn't converge either.

>> Let's think about what happens for some other values of r.

Remember, to evaluate a series I want to to take a limit of the sequence of

partial sums.

These symbols, here's the partial sum,

s sub n is the nth partial sum of this geometric series.

And it's the sum of the terms from k = 0, all the way up to n.

So here's the k = 0 term, here's the k = 1 term.

Dot, dot, dot hides all the other terms.

And then here's the k = n term.

And the real question is what's the limit of these partial sums.

If I take the limit of this as n goes to infinity.

That's the value of this series.

>> I want to compute the limit of that sequence.

>> So I really want to compute the limit of this the nth partial sum.

And if I can compute this limit,

then I've computed the value of the geometric series.

I’ve been calling this nth partial sum s sub n.

So compute the limit of s of n,

first thing to do is get a better handle on s of n.

And the trick for doing that is to multiply s of n by one minus r.

What's that equal to?

Well that's 1- r times, here's an expression for s sub n, right?

The n partial sum.

Just the sum of terms, k from 0, up to k equals n.

But now I can distribute.

So here I've just written down 1- r, times that quantity.

I haven't changed anything but by distributing, I can rewrite this

as 1 times s sub n minus r times sub n.

And now, I can do a little bit more manipulation here.

Subtracting r times this, is the same as subtracting this.

I can multiply each of the terms by r.

But now each of these terms also can be rewritten,

r times r to the 0 is r to the first, r times r to the first is r squared and so

on until I get to r times r to the n which is really r to the n plus one.

So instead of writing it this way, I could rewrite r times s sub n as

minus r to the first plus r squared until I get to r to the n + 1.

But now a wonderful thing happens.

Lots of these terms cancel terms up here.

Well how so?

R to the 0 survives, but

r to the first is killed by this subtract r to the first here.

Inside the dot dot dot there's an r squared.

And that's killed by this r squared term.

And so on, all the terms appear except for the r to the 0,

all of these terms are killed by something down here.

But then I've also got this extra r to the n + 1 term that

doesn't have anything up hear to kill.

So what's the sequel to?

Well this ends up being equal to r to the zero, which survives,

all of the middle stuff dies.

And then this last r to the n + 1 term survives, and I've written it down here.

But r to the zero can be written as just one, so

I could rewrite this as 1 minus r to the n plus one.

All of this is the same that (1- r) times this n partial sum is 1- r to the n + 1.

Let's divide by 1- r.

Okay, so we've got (1- r) times s sub n is 1- r to the n + 1.

And assuming that r is not 1,

then I can divide both sides by 1- r and I get

that 1- r times s sub n over 1- r is equal to 1- r to the n + 1 over 1- r.

Now I can cancel the 1- rs.

And I have got a formula for the nth partial sum.

The nth partial sum is 1- r to the n + 1 over 1- r.

>> Now I will take the limit.

>> So the limit of s sub n, well here is a formula for s sub n.

So the limit of s sub n is just the limit as n goes to infinity

of 1- r to the n + 1 over 1- r.

That's the limit we have to calculate.

>> When does that limit exist?

>> We can compute this limit using our limit laws.

This is a limit of a quotient, so it's the quotient of the limits.

Limit of denominator still needs to be constant.

And limit of a numerator's limit over difference,

which is the difference of the limits.

All that is to say that, using the limit laws, this ends up being 1 minus

the limit as n approaches infinity of r to the n + 1, over 1- r.

Now how do I calculate this limit?

What's the limit of r to the n + 1 as n approaches infinity?

Well the situation is that if r is bigger than one, or r is less than -1,

then the limit of r to the n + 1 as it approaches infinity isn't a finite number.

And consequently, when this happens, when r is bigger than one or

less than minus one, then the limit of the partial sums diverges, and

consequently the original geometric series diverges.

>> But if r is between minus one and one, then we're good.

Exactly.

In that case if r is between -1 and

1 then the limit of r to the n + 1 is n approaches infinity is equal to 0.

If I take a number between -1 and 1 and

raise it to an enormous power that number gets very close to zero.

Well this then lets me calculate this.

If I want to calculate the limit of 1- r to the n

+ 1 over 1- r, that then is 1 over 1- r.

Because the limit of r to the n + 1 is 0.

Lets summarize the situation for geometric series.

So the geometric series, the sum k goes from 0 to infinity of r to the k

is equal to 1 over 1- r, if r is between -1 and 1.

And we know that because we calculated the limit of the partial sums to

be 1 over 1- r.

In the case where r = 1 or r = -1 we already saw it diverged.

And if r is bigger than 1 or r is less than -1 then it also diverges.

So this actually covers all of the cases.

[SOUND]

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