“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

审敛法

在第三个模块中，我们学习用于确定级数是否收敛的各种审敛法：特别地，我们将说明比值审敛法、根值审敛法和积分审敛法。

- Jim Fowler, PhDProfessor

Mathematics

Let's integrate.

[MUSIC]

Series and integrals are not entirely unrelated.

Here I've got a graph, and I've drawn a bunch of boxes in red on my graph.

Each of these boxes have width one, and

the heights of these boxes given by the terms in a sequence.

So the first box has high a sub one.

The next box has high a sub two.

The next box has high a sub three.

The next box has high a sub four, and so on.

Now since the boxes each have width one, it's easy to calculate their areas.

The area of the first box is one times its height, which is a sub one.

So the area of the first box is a sub one.

The area of the second box is its width one times its height a2 sub two.

So the area of the next box is a sub two and so on.

The area of the next box is a sub three.

The area of the next box is a sub four, or

if I want to calculate the area of all the red boxes.

I imagine these red boxes go on forever.

Well that's really an infinite series right?

It's a sub one, plus a sub two, plus a sub three, plus a sub four and so on.

So I could write the total area in all the red boxes as this infinite series.

There's more going on in this graph.

I don't just have these boxes.

I've also got this blue curve.

That's the curve y = f(x).

And one special feature about this function f is that I've rigged it,

so that a sub n is f(n).

And we can see that here because this blue curve passes through

the corners of these boxes.

Now, what if I wanted to calculate the area under the blue curve and

to the right of the line x equals one?

Well, that would be this integral.

The integral from one to infinity of f(x)dx.

How does this blue region compare to this red region?

Well this blue region sits inside all of the red boxes.

What does that mean?

That means that this is the red region this is this blue region,

but this blue region sits inside all of the red boxes.

And that means that the area of the red region must be larger

than the area of the blue region.

Why does that matter though?

Well the take away message here,

is that this sum is even larger than this interval.

And that means, if I know that this integral, say is infinity.

If the area under the blue curve and to the right of the line x = 1 is infinite.

Then what do I know about this series?

Well that series then, must diverge.

The picture includes some as of yet unspecified assumptions.

So more precisely, I need to assume that f is a decreasing and positive function.

And I should explicitly mention that an = f(n).

And with those conditions, then if this integral diverges.

Then the sum also diverges.

We're in a position to prove this.

This what I'd like to show.

Suppose that little f is a decrease in the positive function

building a sequence a sub n out of the function f.

And supposing that the integral f of x dx goes from one to infinity diverges,

and I want to show that this is series diverges as well.

We're going to get a handle on the series, I'm going to look at this.

The sum little n goes from one to big N of a sub n.

And somehow, I want to relate this sum to an integral.

Well I won't get that all at once, but I'll do it in steps.

First thing I can say is this.

That the sum of the a sub n's is equal to the sum, little n goes from

one to big N of the integral x goes from n to n+1 of f of n dx.

And although this looks somewhat fearsome.

It's true for a not very deep reason.

The upshot here is what, f(n) is equal to a sub n.

And consequently,

this is just the integral of a constant on interval of length one.

So if you integrate a sub n.

X goes from n to n plus one.

Well, you're just getting a sub n.

So this is the sum of a sub n, which is exactly what I got here.

But when I write it like this.

I'm giving myself some more freedom to mess around with this integrate.

Instead of saying f(x).

I could say f of the floor of x.

Why is that helpful?

Well, what's floor of x?

All right.

The floor of x is the greatest integer less than or equal to x.

So if x is between n and

n + 1 as an integer then the floor of x is just equal to n.

And that means except at the end points,

f to the floor of x is equal to f of n on the interval of n and n plus one.

And that means integral is just again a sub n.

So the sum of the a sub n's is the sum of, well that's just a sub n.

Now I'm writing a sum of a bunch of integrals.

All the integrands look the same, but the bounds of integration are changing, right?

The first integral goes from one to two.

The next integral goes from two to three.

The next integral goes from three to four, and so on.

But if you add up a bunch of integrals, all of which lie end-to-end like that.

You can just write that as a single integral.

So this sum of integrals is equal to just the integral x goes from

one to N + 1 of f of the floor of x dx.

All right, so I've replaced this sum with this integral.

Now what can I do.

Well I know that f is a decreasing function.

So if I plug in a bigger input the functions output is smaller.

And that tells me the that the integral of f of floor Lx.

x goes from 1 to big n + 1 Is bigger than or

equal to the integral of just f of x, x goes from 1 to big N plus 1.

Because f of the floor of x, is at least as big of f of x.

Now what else do I know?

I know that this integral, diverges.

Which means what?

Well, It means that as long as N is big enough.

I can make this as big as you'd like.

All right.

All you need to do is tell me how big you want this to be, and

I'll tell you some big enough N.

So that this thing is really that large.

But, this thing is smaller than this sum.

So what that means is that this sum

can be made as big as you like as you choose big and big enough.

But if you can make this sum really large

that means the sequence of partial sums isn't getting close to any finite value.

And what that means is that this series diverges.

We can go the other way too.

We can integrate to prove convergence.

So again, I've got this interval and this series, and I somehow want to relate them.

Well, let's suppose that this interval is finite.

Then I'd like to be able to conclude that this series converges.

Well all the same conditions as before, right?

I want f to be a positive decreasing function and such.

Let's prove it!

So here's what I've got.

I've got that this integral is finite, converges.

Then I want to know that the series converges as well.

So here's the integral, the integral from one to infinity f(x) dx.

How does that integral compare to just the integral from 1 to N of f(x) dx?

Well, f's a positive function, so these things differ by the integral from

N to infinity, which is some positive number.

So this thing here is definitely an overestimate of this thing.

So the integral from one to infinity is bigger than the integral

from one to big N of f of x dx.

Now I'm going to relate that integral to this integral.

The integral from one to big N of f of the ceiling of x.

Remember what the ceiling is.

The ceiling is the smallest integer bigger than or equal to x.

So how does f of x relate to f of the ceiling of x?

Well the ceiling of x is always at least as big as x.

But because f is decreasing f sends bigger inputs to smaller outputs.

So f with a ceiling of x is less than or equal to f of x.

And that means this integral is an overestimate of this integral.

Now, I can split this integral up.

Instead of integrating from one to big N all in one fell swoop

I can instead integrate like this.

I can integrate from little n to little plus one,

and add up all those integrals as n goes from one to big N minus one.

But on each of these intervals, I know exactly what the ceiling of x is.

If x is between n and n+1,

then the ceiling of x is just n+1.

So, now I've got the sum of the integral of f of n+1, but what's f of n+1?

Remember f of n plus one is just a sub n plus one.

And I'm integrating this constant over an interval of length one.

So this integral just ends up being a sub n plus one, and

that means these are equal.

Now, I could re-index this.

The first term that I'm adding here, the n equals one term is a sub two.

The next term is a sub three.

And eventually, I plug in the little n equals big N minus one,

which gives me a sub big N here.

So I could instead right that as just the sum,

little n equals two to big N of a sub n.

So all together, what I've shown is that this integral, which has some

finite value, is an overestimate of this, which is almost the partial sum.

It's just missing a sub one.

What that means is that the sequence of partial sums for this series, is bounded.

And I already know that it's monotone,

because all the things that I'm adding up are positive.

So, I've got a sequence which is bounded and monotone and

therefore, by the monotone convergence theorem.

This sequence of partial sums converges which exactly what it means,

this series converges.

We're doing one more than just proving convergence.

What I mean is that I've got this inequality here.

The big deal here is that I can add a sub one to both sides, and

if I add a sub one to both sides.

Now I've got this inequality.

But this is just a sub one, plus a sub two, plus a sub three, plus dot,

dot, dot, plus a sub big N, which I all ready know is least as

large as the integral from one to big N plus one at f(x) dx.

Let's now take a look at this.

I could now take a limit as big n approaches infinity and

here I would be getting what?

Well, this would just become the series, the sum of all the a sub ns.

And this is now the integral from one to infinity of f(x) dx.

So what I really know is that the value of this series is trapped

between this integral, and the same integral, plus the first term.

Let me summarize what we have.

Same setup as always.

I'm supposed that I've got some function, little f which is decreasing but positive,

and I built my sequence out of that function.

Then, if I've got that this integral converges.

If that integral has a finite value,

that happens if and only if this series converges.

Right, because you've shown both directions.

I've shown that if this integral diverges, then this series diverges.

But I've also shown that if this integral has a finite value.

Then the sequence of partial sum series is monotone and bounded.

Therefore, this series converges.

And more than that, I've also got these bounds.

I also note the value of this sequence is trapped between

these two values.

[SOUND]

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