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Let's integrate.

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Series and integrals are not entirely unrelated.

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

Each of these boxes has width one and the heights

of the boxes, given by the terms in a sequence.

So the first box has height a sub 1, the next box has height a sub 2, the

next box has height a sub 3, the next box has height a sub 4 and so on.

Since the boxes each have width 1, it's easy to calculate their areas.

The area of the first box is 1 times it's height, which is a sub 1.

So the area of the first is a sub 1.

The area of the second box is it's width 1 times it's height

a sub 2, so the area of the next box is a sub 2.

And so on.

The area of the next box is a sub 3, the area of the next box is a sub 4.

What if I want to calculate the area of all

the red boxes, I imagine this red boxes go on forever.

Well, that's really an infinite series, right?

Its a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 and

so on, so I could write the total area

in all the red boxes as these infinite series.

1:15

There is 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 equals f of x and

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

that a stubborn is f of n and we can see that here.

Because this blue curve passes through the corners of these boxes.

1:38

Now what if I wanted to calculate the area under the

blue curve and to the right of the line x equals 1?

Well that would be this interval.

The interval from one to infinity of f of x, dx.

How does this blue region compare to this red region?

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

2:15

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 takeaway 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 equals 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 a sub n is equal to f of n and with those

conditions, than if this integral diverges, then the sum also diverges.

We're in a position to prove this.

This is what I'd like to show. Suppose that little f is a decreasing but

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

And I'm supposing that the integral f of x dx goes from 1

to infinity diverges and I want to show that the series diverges as well.

3:34

Well to get a handle on a series, I'm going to look at this.

The sum little n goes from 1 to big N the a

sub n, and somehow, I want to relate this sum to an interval.

I won't get there 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 1 to big N of the integral x goes from n to n plus 1 of f of n dx.

And although this looks somewhat fearsome It's true for a not very deep reason.

the upside here is what?

F of n is equal to a sub n,

and consequently, this is just the integral of a constant

on an interval of length 1.

So if you integrate a sub n, x goes from n

to n plus 1, well, you're just getting a sub n.

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

4:35

Well why's that

helpful well what's floor of x.

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

So if x is between n and n plus 1, n's an integer, then the floor of

x is just equal to n and that means, except at the endpoints, f of the floor

of x is equal to f of n on the interval between n and n plus 1.

And that means this 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.

5:09

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 1 to 2, the next integral goes from

2 to 3, the next integral goes from 3 to 4 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.

5:51

Oh, I know that f is a decreasing function, so if

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

That tells me that the integral of f of floor of x, x goes from 1 to big N plus 1.

Is bigger than or equal to the integral cause f of x, x goes from

1 to big N plus 1, because f of the floor of x is at least as big as f of x.

6:53

But if you can make the sum really, 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 integral and this series and I somehow want to relate them.

Well let's suppose that this integral is finite.

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

With 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, that I

want to know that the series converges as well.

So here's the integral.

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

How does that integral compare just the integral from one to big N of f of x dx.

7:47

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 over estimate of this thing.

So the integral from

1 to infinity is bigger than the integral from 1 to big N of f of x, dx.

I'm going to relate that integral to this integral, the integral from

1 to big N of f of the ceiling of x.

Remember what the ceiling is, the ceiling is

the smallest integer bigger than equal to x.

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

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

But because f is

decreasing, f sends bigger inputs to smaller outputs.

So f for the ceiling of x is less than or equal to f of x.

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

Now I can split this integral up and instead of integrating from 1

to big N all in one fell swoop, I can instead integrate like this.

I can integrate from little n to little n plus 1 and

add up all those integrals as n goes from 1 to big n minus 1.

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

If x is between little n and little n plus 1, then the ceiling of x is just n plus 1.

So now I've got the sum of the integral of f of n plus 1.

But what is f of n plus 1?

Remember f of n plus 1 is just a sub n plus 1.

And I'm integrating this constant of an interval of length 1, so this

integral just ends up being a sub n plus 1 and that means these are equal.

Now I could re-index this, you know, the first

time that I'm adding here, the n equals 1 term,

is a sub 2, the next term is a sub 3, and eventually I plug in a little n equals

big N minus 1.

Which gives me a sub big N here, so I could instead write that

as just the sum little n equals 2 to big N of a sub n.

9:49

So all together, right, what I have 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 1,

but what that means is of a 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 conversions theoreom,

this sequence of partial sums converges, which is exactly

what it means to say that this series converges.

We're doing more than just proving convergence.

What I mean is that I've got

this inequality here.

The big deal here is that I could add a sub 1 to both sides

and if I add a sub 1 to both sides, now I've got this inequality.

But this is just a sub 1 plus a sub 2 plus a sub

3 plus dot dot dot plus a sub big N, which I already know.

10:49

Is at least as large as the integral from 1 to big N plus 1 of f of 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 n's. And this is now the integral from 1 to

infinity of f of x dx. So 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 supposing I've got some function little f which

is decreasing but positive and I built my sequence out of that function.

11:38

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 the series diverges.

But I've also shown that if this interval has a finite value, then

the sequence of partial sums is

monitored and bounded therefore the series converges.

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

I also

know that the value of this sequence is

trapped between these two values.

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