“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。

Loading...

From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

46 ratings

“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 compute the value of another series.

[SOUND] A bit of algebra will

make a huge difference here.

1 divided by k + 1 times k, is the same as 1 over

k- 1 over k + 1, see why this is true, right?

I can write these over a common denominator of k + 1 times k,

that means that 1 over k is k + 1 over k + 1 times k and

1 over k + 1 is k over k + 1 times k.

So, now I've got k + 1- k in the numerator and that's just 1, why is that helpful?

It will take us a little while to see how that algebra fact helps, but

in the meantime remember what our goal is.

Our goal is to evaluate this series, and

instead of just diving straight into that, let's evaluate a slightly simpler sum.

Instead of k goes from 1 to infinity,

let's just do k goes from 1 to 5 of the same thing.

Of course, this is small enough that we can just do the calculation, so

to calculate this, I just plug in k = 1, k = 2, k = 3, k = 4,

k = 5 into this expression and add it all up.

Let's see, I plug in k = 1 and I get 1 over 2 times 1, and

I'll add that to what I get when I plug in k=2, which is 1/3 times 2.

And I'll add that to k=3 term, which is 1/4 times 3.

And I'll add that to the k=4 term, which is 1/5 times 4.

And I'll add that to the k=5 term, which is 1/6 times 5.

So all I got to do is do this arithmetic, and I can simplify this a bit,

instead of write 1 over 2 times 1 I’ll just write a half,

I got 1 over 3 times 2 is a 6th, here I got 1 over 4 times 3 that’s a 12th,

here I got 1 over 5 times 4 that’s 20th, here I got 1 over 6 times 5 that’s a 30th.

All I've got to do is add up these fractions.

What I can do, I put them over common denominator of 60.

Here we go, one-half in sixtieth is thirty-sixtieth.

One-sixth in sixtieth is ten-sixtieth.

Over common denominator 60, this becomes five-sixtieth,

common denominator is 60, this becomes 3 60ths.

Common denominator of 60, this becomes 2 60ths.

And just add up the numerators,

30 plus 10 is 40 plus 5 is 45 plus 3 is 48 plus 2 is 50.

So this entire thing is just 50 60ths.

But instead of writing 50/60, I could just write 5/6.

Alternatively, we can make use of the algebra fact from the beginning.

What was that algebra fact?

It was the fact that 1/(k+1) times k is the same as 1/k- 1/k+1.

Now, why is that helpful?

Well that means, instead of writing down 1 over k + 1 times k, I can write down this.

So when k = 1, instead of writing down this, I'd write down this.

1 over 1- 1 over 2.

And when k = 2, instead of writing down this, I'd write down this.

1/2- 1/3 and when k = 3,

instead of writing down 1 over 4 times 3 I'd write down this.

1/3- 1/4.

And when k = 4, in stead of writing this, 1 over k plus 1 times k,

I'm going to write down 1/4- 1/5 and the last term the k = 5 term.

Instead of 1 over 6 times 5, it'll be 1 over 5- 1 over 6.

Why is this an improvement?

Well the cool thing that happens here is that most of these terms end up canceling.

Take a look, I've got a minus 1/2 plus 1/2, minus 1/3 plus 1/3,

minus 1/4 plus 1/4, minus 1/5 plus 1/5.

The only terms that survive here are this initial 1 over 1 term and

this last 1 over 6 term.

And that means, this entire thing ends up just being equal

to 1 over 1 minus 1 over 6 which is 5 sixth.

This trick has a name.

We see that the series telescopes.

And we'll call such a thing a telescoping series.

What about the original series, with infinitely many terms?

This is the original series that I want to compute the value of.

And this is by definition equal to the limit as n approaches infinity

of the sum k goes from 1 to n of 1 over k + 1 times k.

We can compute this limit.

Well this limit is the limit as n approaches infinity

of the sum k goes to n, and instead of writing this,

all right, one over k minus one over k plus one.

And this will be a much better way to write it, right?

Because what is this, this is then the limit as n approached infinity

of what we write in the first term, it's one over one minus one over two.

Plus the next term, the k equals 2 term,

which is plus one half minus a third, and I'll write dot, dot, dot,

plus the last term, which is 1 over n minus 1 over n plus 1.

That's when k equals n.

But practically all of these terms die.

This minus one half is killed by this one half.

This minus a third is killed by something.

Every single term here.

There's a term in here which kills this 1 over n.

The only thing that's left is this initial 1 over 1 and this final -1 over n + 1.

So this limit is just the limit as n goes to infinity

of 1 over 1- 1 over n + 1.

But what is that limit?

Well, this is 1 minus 1 over an enormous number, right?

This limit is just 1.

Let's formulate this as a general trick.

Suppose that I want to take a sum, k goes from 1 to n,

of some function evaluated at k, minus the same function evaluated at k+1.

Then this is what, well it's f of one minus f of

two plus f of two minus f of three, and so

on until finally I get to the last we'll then,

it's f of n minus f of and then plus one.

And just like before, practically all the terms cancel.

This F(2) and this F(2) cancel.

This -F(3) cancels with something and the previous term here cancels this F(n).

So this sum is equal to F(1) -F(

Of n plus 1, and taking a limit let's us say something about the infinite series.

So now suppose I want to calculate the sum

k goes from 1 to infinity of (f (k) minus f (k plus 1)).

Well, this infinite series is by definition just the limit as

n approaches infinity of the sum k goes from 1 to n

of f of k minus f of k plus 1.

But, I just saw how to calculate this.

This is now the limit as n approaches infinity of a, what's this?

This is f of 1 minus f of n plus 1.

But this is now a limit of a difference, and

the limit of this first term's the limit of a constant which is f of 1.

So this is equal to f of 1 minus

the limit As n approaches infinity of the function f.

Let's do another example.

This fact again.

Let's try it.

Let's try to evaluate the sum k goes from 1 to infinity

of k over k plus 1 factorial.

And the trick is to rewrite this as telescoping sum, so

I gotta cook up some function, and

the function I'm going to use with a function F of K is one over K factorial.

And I just want to check that F of K minus F of K plus 1 what is that?

That's 1 over K factorial minus 1 over

K plus 1 factorial, but right isn't a common denominator of K plus 1 factorial.

This is K plus 1 over K plus 1

factorial minus 1 over (k+1)!.

And what happens here is that I've got a (k+1)!

in the denominator and a k+1-1 in the numerator, that's a k.

So indeed, this sum is

the sum k goes from 1 to infinity of this function.

One over k factorial minus this function at k plus one which

is one over k plus one factorial.

But now, this infinite series is just it's value when this function

of k equals one which is one over one factorial minus the limit.

As n approaches infinity of this function.

I'll write 1/(n+1)!.

But this limit is 0.

And that means that this infinite series has value 1, and it.

[SOUND]

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.