Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

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Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Series

In this second module, we introduce the second main topic of study: series. Intuitively, a "series" is what you get when you add up the terms of a sequence, in the order that they are presented. A key example is a "geometric series" like the sum of one-half, one-fourth, one-eighth, one-sixteenth, and so on.
We'll be focusing on series for the rest of the course, so if you find things confusing, there is a lot of time to catch up. Let me also warn you that the material may feel rather abstract. If you ever feel lost, let me reassure you by pointing out that the next module will present additional concrete examples.

- Jim Fowler, PhDProfessor

Mathematics

We need general techniques for determining when a series converges.

[MUSIC].

We've already seen an example of what's usually called the Comparison Test.

The original series that we looked at was this series.

It has the sum, n goes from 0 to infinity.

Sine squared n, divided by 2 to the n. And that series converges.

But, remember why.

Let's recall how we prove that this series converges.

And what we did was to compare it to a geometric series.

So, 0 is less than or equal to sine

squared n over 2 to the n, is less than or equal to 1 over 2 to the n.

And these inner qualities just follow from the fact

that sine squared is trapped between 0 and 1.

And what do I know about the sum of 1 over 2 to the n?

The sum 1 over 2 to the n, n goes from 0 to infinity, converges.

Right, this is a geometric series, I can even evaluate it.

Now, what does

that tell me about the original series that I'm studying?

What about the sine squared n over 2 to the n?

Well the sequence or partial sums for this series are

all bounded above by the vast value of this series.

And the sequence of partial sums is non

decreasing because these terms are all none negative.

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

That means the sequence of partial sums converges, and that's just what it means

to say that this series converges. That was a bit quick.

So let's generalize this argument, try to

formulate it as a broadly useful technique.

So here's the usual setup.

I'm, imagine that I've got two series. I'll use K as the index, K goes from 0 to

infinity of a sub k, and another series, K goes from 0 to infinity of b sub k.

And I'm supposing that 0 is less than or equal

to a sub k, is less than or equal to b sub k, for all K.

I'm imagining that the sum of the b sub k's converges.

Right.

So I'm going to be assuming that this series,

converges, and then I'm wondering about this series.

Does that series converge, or diverge? Well, to study that question, supposed to

look at the sequence of partial sums. So that's S sub n is the sum

of the terms, K goes from 0 to n, of a sub k.

And now, the whole question is whether this sequence converges?

Because the convergence of this sequence is exactly

what it means to say that this series converges.

What kind of sequence is that? Well, here's what I know.

I know that a sub k is less than or equal to b sub k.

And that means if I add up a whole bunch of a sub k's.

Say, the a sub k is between k equals 0 and n.

That's less than or equal to adding up the b sub k's, K goes from 0 to n.

Now this thing here is just the nth partial sum.

Now what do I know about this sum?

Well the rest of the b sub k's are all non negative, so if I add up even more b sub

k's, I'm just making this even bigger. So what I can conclude,

is this. That s sub n is less than or

equal to the value of the series, K goes from 0 to

infinity, of b sub k. In other words, what I can conclude is

that S sub n, is bounded above. And it's monotone.

So let's check that.

I'm going to show that if m is bigger than n, then

S sub m is at least as big as S sub n.

Well, what's S sub m?

S sub m is the sum, K goes from zero to m, of a sub k.

And I'm sure that's bigger than or equal to the sum K goes from 0 to n, a sub k.

But, m is bigger than n,

so I can re-write this sum. I can split this sum up.

This is the sum K goes from 0 to n, of

a sub k, plus the terms between n plus 1 and m.

So, K goes from n plus 1 to m of a sub k, and that's at least as big, I hope,

as the sum K goes from 0 to n of a sub k. Now, why is this true?

Well,

I've got the sum K goes from 0 to n of a

sub k on both sides, but I've got this extra term over here.

What I need to know is that this extra term is at least 0.

But that's true, because all the a sub k's are non negative, and if you

add up a whole bunch of non negative

numbers, well, then, the result is non negative.

So this is telling me that the sequence of partial sums, is non decreasing.

So the

sequences of partial sums is non decreasing and bounded above.

So with a monotone convergence theorem, it converges.

Let's summarize that.

So here's what we've proved. If we've got two sequences, which are

related in this way, that a sub k is between 0 and b sub k, and the sum, K goes

from 0 to infinity, of b sub k, converges. Then,

the sum K goes from 0 to infinity of a sub k, also converges.

We can also say something when the sum of the a sub k's, diverges.

Suppose that 0 is less than or equal to a sub k, is less than or equal to b sub k.

And let's suppose that the sum of the a

sub k's, K goes from 0 to infinity, diverges.

And I don't know, what can I say about the sum of the b sub k's?

Well, if this series diverges, that means that the sequence of partial sums,

the sum, K goes from 0 to n of a sub k. That can't converge,

right, because to say this thing converges is to say the

series converges, so if this series diverges, this sequence must diverge.

But that sequence, is monotone for the same reason as before, right.

I'm adding up non negative terms, so as I

add up more terms, at least is non decreasing.

So the sequence is a monotone sequence, and yet, that sequence doesn't converge.

So that means that the sequence of partial sums must be

unbounded, because if it were bounded, and it was

monotone, then it would converge, but it doesn't converge.

So I've got an unbounded sequence.

What about the sequence of partial sums for

the series involving the b sub k's, right?

So the sequence of partial sums there as the sum K goes from 0 to n of b sub k.

This is even bigger than the sequence

of partial sums for the a sub k's. This sequence, is likewise unbounded.

Well, if this sequence is unbounded then that sequence doesn't converge.

And if that sequence doesn't converge that means

that the sum of the b sub k's, diverges.

And that was fast.

So we hope to see the convergence and divergence statements together.

Well, here's our standing assumption. That 0 is less than or

equal to a sub k, is less than or equal to b sub k.

And the first thing that we saw, is that if the series of

the b sub k's converges, then the series of the a sub k's converges.

On the other hand, if the series of the a sub

k's diverges, then the series of the b sub k's diverges.

It's super important to keep track of the directions of these implications.

Alright, here I've got the convergence

of b sub k implying the convergence of a sub k.

And here I've got the divergence of a sub k implying the divergence of b sub k.

And hopefully that makes sense, right?

This is saying that as long as you've got non

negative terms, if you're below a convergent series, you converge.

And this is saying, as long as you've got non

negative terms, if you're above a divergent series, you also diverge.

And let me end by saying

the Comparison Test is just a ton of fun to use.

It's definitely my favorite test.

[SOUND].

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