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

Convergence Tests

In this third module, we study various convergence tests to determine whether or not a series converges: in particular, we will consider the ratio test, the root test, and the integral test.

- Jim Fowler, PhDProfessor

Mathematics

Let's look at ratios.

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We've already seen how to determine the convergence of a geometric series.

For example, does the series the sum n goes from 1 to infinity

of 1 over 4 to the n converge or

diverge? Well this series converges, right.

By which test? By the geometric series

test, if you like.

Because the common ratio, all right, this common ratio between

the terms is 1 4th And that's less than 1.

What if we modify the series somewhat? So what if instead of looking at this

series, we looked at the series the sum n goes from 1 to infinity out of 1

over 4 to the n, but say n to the 5th over 4 to the n.

That series is not a geometric series anymore, but

it ends up at least approximately looking like one.

Let me just give some names to these things.

Let's say that a sub n is n to the 5th over 4 to the n.

And what I'm trying to determine is the sum n goes from 1 to infinity of a sub n.

And I'm probably not going to actually determine this value right.

But I'm just trying to answer

the question of whether or not that series converges or diverges.

Now, if I plug in really big values for n,

I can get a sense of what these terms look like.

All right.

A sub 1000 is well, hard to write down, but

I can at least write down 1000 to the 5th.

Now if this were actually a geometric series, right,

the ratio between subsequent terms would be the same.

This isn't a geometric series, but let's see if we're close.

So a sub 1001 divided by a sub 1000.

Well a sub 1001 is 1001 of the 5th power over 4 to the 1001st

power divided by a sub 1000, which is 1000 to the 5th power divided

by 4 to the 1000th power. I could rewrite this, right?

because I've got a fraction with fractions in the numerator and the denominator.

I could rewrite this as 1001 to the 5th times 4

to the 1000th power, divided by 1000 to the 5th times.

And the denominator and the numerator, write that

in the denominator, is 4 to the 1001st power.

Now I should split this up and think about these two pieces separately.

Right?

How big are, are these two pieces?

Well, 1001 to the 5th power divided by 1000 to

the 5th power, this piece here is pretty close to 1.

And this piece here, is, well it's exactly 1 4th, so if I'm got a number

that's close to 1 and I'm multiplying it by a number that's close to 1 4th,

this fraction is well, it's not equal to, but it's about 1 4th.

And if you think about it, there's nothing

special about 1000 here, except that it's really big.

So the ratio between subsequent terms, at least if those those terms are

far enough out in the sequence, that ratio is close to 1 4th.

Can I make this precise? I'll use a limit,

what I'm saying is that the limit as n approaches infinity of a

sub n plus 1 over a sub n, in this case is equal to a quarter.

That if I choose n big enough, I can make that ratio close.

How close? As close as you want to a quarter.

I can be very explicit in this case.

I, I could pick an explicit value of epsilon.

Let's just

pick 0.1.

And then it turns out that whenever little n is bigger than 15, a sub n plus 1

over a sub n is, in fact, within epsilon of my limiting value of a quarter.

And in particular if a sub n plus 1 over a sub n, is within 0.1 of a quarter.

Well that also means that a sub n plus 1 over a sub n is

less than a quarter plus 0.1.

And that's enough to set up a comparison with a geometric series.

I could multiply this inequality by the positive number a sub n.

And I'd find out that a sub n plus 1 is less than a quarter plus

0.1 times a sub n. Well, this also means that a sub n plus 2

is less than a quarter plus 0.1 times a sub n plus 1.

But a sub n plus 1 is less than a quarter plus 0.1 of a sub n,

so a sub n plus 2 is less than a quarter plus 0.1 squared

times a sub n. Now, this fact with n replaced

by n plus 2, tells me that a sub n plus 3 is

less than a quarter plus 0.1 times a sub n plus 2.

But a sub n plus 2 is less than a quarter plus 0.1 squared times a sub n,

so a sub n plus 3 is less than a quarter plus 0.1 cubed times a sub n.

Well, this works in general, right.

What I'm really showing here is, is what? I'm showing that a sub n

plus k is less than a quarter plus 0.1 to the kth power

times a sub n, and this is true whenever n is at least 15.

So in particular, I'm showing that a sub 15 plus k.

Is less than a quarter plus point 1 to the kth power times a sub 15.

The key

point here is that a quarter plus epsilon is still less than 1.

Well, the series that we're interested in is the sum

n goes from 1 to infinity of a sub n.

And that series converges precisely when the series n

goes from 15 to infinity of a sub n converges.

The first 14 terms here that aren't included there don't effect convergence

at all. All right?

That's just adding 14 more numbers to this.

But now how do we analyze this?

Well here's the deal right, I've got a bound on the size of these

a sub n's as long as n is at least 15, I got this.

So the sum n goes from 15 to infinity of a sub n

is equal to the sum k goes from 0 to infinity of a sub 15 plus k.

Right, this 0 term here is just a sub 15,

which is exactly the same as the first term here.

K equals 1 is a sub 16 here, and that's exactly the next term here.

So these are really the same series.

But I, what do I know about this series? Well that series converges, why?

Because the sum k goes from 0 to infinity of a

quarter plus 0.1 to the k times a sub 15 converges.

Why does this series convergence affect this series convergence?

Well, this series's terms are bigger than this series terms.

In this series, terms are all positive.

So this follows from the comparison test, that

if this series converges, then this series converges.

But now why does this series converge?

Well, this series converges by, say, the geometric

series test.

And how does the geometric series test effect this?

Well, this is a geometric series. This a sub 15 is just a constant, and the

common ratio is a quarter plus 0.1, and a quarter plus 0.1 is less than 1.

So the common ratio, being less than

1, means that this geometric series converges.

Means that this series converges.

Means that the original series that we're

studying converges.

This is a specific case of a very general procedure.

That I'm studying the convergence of some series the sum

n goes from 1 to infinity of a sub n.

All right, I want to know if that converges or diverges, and let's

suppose that all the terms in the sequence a sub n are positive.

What I'm suggesting that we try to do is to look at the limit as

n approaches infinity of a sub n plus 1 over a sub n.

All right.

This series is almost surely not a geometric series, right?

Most series aren't geometric series, but I could try

to see, is it close to a geometric series?

What is the ratio between subsequent terms?

Is it close to anything eventually, right?

That's what the limit's asking me.

So let's suppose that this limit does exist, let's suppose that

this limit equals L, and let's suppose that L is less

than one.

Well, then what I'm suggesting is that we pick some small epsilon.

how small?

Well, I went epsilon so small that L plus epsilon is less than 1.

If L is less than 1, I can definitely pick epsilon small enough,

so that even if I add epsilon to L, I'm still less than 1.

I'm going to use epsilon in this limit, all right.

The fact that this limit equals

L, means that there's some big N, let's just say, find big N.

So that whenever little n is bigger than or equal to big N, a

sub n plus 1 over a sub n, should be within epsilon of L.

And, in particular, this should be less than L plus epsilon.

And why does that help? Well, this inequality implies

that a sub big N plus little k is less than L plus

epsilon to little k times a sub big N. Right, where does this come from?

Well it really does come from just applying this a ton of times.

What this is saying is that as long as n is at least big N, then the ratio between

subsequent terms is less than this factor. And that means that I can

overestimate the big N plus kth term by this factor,

the kth power times, just whatever the big Nth term is.

Now, how do I use this?

Well, the sum, k goes from 0 to infinity of this, of L plus epsilon to the k

times a sub N converges, and why is that? Well this is a geometric series with

common ratio L plus epsilon. But that common ratio is less than 1.

All right, that's exactly why I picked epsilon so small up here.

So I've got a geometric series with common

ratio less than 1, so this series converges.

But then, by the comparison test, I know something about this series.

I know that the series little n equals big

N to infinity of a sub n also converges

because these terms are all over estimated by these terms.

And consequently, the original series, the sum n goes from 1 to

infinity of a sub n, also converges. Because I

can get this from this, just by adding on finitely many extra terms.

We can summarize this.

So here's a statement of what's normally called the ratio test.

And it goes like this.

I want to know whether the series, n goes from one to infinity of a sub n,

converges or diverges. And I'm only going to study series,

the terms of which are all positive. And I need that because my argument

was using the comparison test, and the comparison

test requires series with at least non-negative terms.

All right.

Then I'm going to look at the limit n goes to infinity of a sub

n plus 1 over a sub n, the limit of the ratio of subsequent terms.

Let's suppose that that limit exists

and equals L, then there's three possibilities.

If that limit is less than 1,

then this series converges and it's converging because I can do a

comparison test between this series and a convergent geometric series.

If that limit is bigger than 1, then this series diverges.

And I know that because, again, if this limit is

bigger than 1, I can compare, maybe not the whole series.

But at least after a while, for some, past some big nth term, I can compare this

series to a divergent geometric series. And then finally, maybe L is equal

to 1.

And in that case, the ratio test is silent.

It doesn't tell us any information either way.

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