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The ratio test.

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The ratio test asks us to look at the limit of the ratio of subsequent terms.

Well, here's a precise statement.

So the question is whether the sum n goes from

one to infinity of a sub n converges or diverges.

I'm going to want to assume that all of the terms that I'm adding up, all of

the a sub n's, are positive, in order that I can use the comparison tests.

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And maybe that limit is equal to one.

In which case, the limit test is simply silent.

It might converge, it might diverge.

We just don't know.

The ratio test doesn't tell us any information when L is equal to one.

Why does this work?

What if the limit is less than one? So I'm trying to figure out what happens

when big L is less than one.

I want to show the series in that case converges.

Well, if big L is less than one, I can pick some very tiny

value of epsilon so that even big L plus epsilon is less than one.

Then I can use epsilon in the definition of limit.

So there must be some big N so that whenever little n is at least

big N, the thing I'm taking the limit of, a sub n plus one over a

sub n must be within epsilon of L.

So in particular, a sub n plus one over a sub n must be less than L plus epsilon.

Now how is this fact helpful? Well, I'm trying to calculate this series.

I'm trying to sum a-sub-n, n goes from one to infinity.

But I can break it up into two pieces, when n is

smaller than big N, and when n is bigger than big N.

Well, here's the piece when n is smaller

than big N.

It's just the sum of you know, a finite sum.

I'm not going to have any trouble adding those numbers up.

But then I've got the rest of the series that I have to add up.

Right, I have to add up a sub big N, plus a sub big N plus one, and so on.

And this is the hard part, right?

Because I've got dot dot dot, forever. But what I know now is that I can

control the ratio between subsequent terms, maybe not here, but at least here.

I know that a sub big N plus one is less than R times a sub big N.

And I know that a sub big N plus two is less than r times a sub big

N plus one which itself is less than r times a sub big N.

So, all of that is to say that this thing here is well, not equal to this,

but this is an overestimate of what was there before.

So now I've got this series of at least

less of than this first part, plus the rest.

But what do I know about the rest?

Well the rest of this thing is now a geometric series and

I know this geometric series converges because little r is less than one.

So I can replace that with this convergent geometric series.

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And in light of this, all right, this is really now setting up a comparison test.

I am really saying that the sequence of partial sums is bounded by this.

And since the sequence of partial sums is a monotone, because all the terms

that I'm adding up are positive, I then know that this series converges.

And what if the limit is bigger than one? So I am trying to figure

out what happens when big O is at least one and

I want to show that in that case the series diverges, or big

O is bigger than one then I can choose some tiny

epsilon so that big O minus epsilon is also bigger than one.

Then I can use that epsilon in the definition of limit.

So there's some big N so that whenever little n is bigger

than or equal to big N, this ratio, the thing I'm taking

the limit of, is within epsilon of the limit L.

So in particular, this ratio is bigger than big L minus epsilon.

Why is that significant?

What I'm trying to calculate is this series.

Or at least I'm trying to figure out if it converges or diverges.

And I can start just adding up terms, right?

A sub one plus a sub two.

Eventually I get up to a sub big N minus one,

then a sub big N plus a sub big N plus one,

and so on.

I keep on adding up terms, making a

limit of partial sums, you know, strictly speaking.

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But I know that all of these terms are positive.

So, if I throw away a bunch of terms, well, this isn't equal to this anymore,

that this is now larger than this thing

where I've thrown away a whole bunch of terms.

We're going to start with a sub big N term.

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Now, what else do I know?

Because a sub

little n plus one over a sub n.

Is bigger than this whenever little n is bigger than or equal to big N.

I can underestimate this.

Alright, this is saying that a sub big N plus

one is larger than r times a sub big N.

Telling me that this, is bigger than r times this.

Which is also bigger than r times this.

Which means this is larger than r squared times this.

This term here, a sub big N plus three is larger than r cubed times this.

What I'm really saying is that this

infinite series is at least this infinite series.

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But what kind of infinite series is this?

Well this is just a geometric series, right?

That's the geometric series.

The sum k goes from zero to infinity of r to the k times a sub big n.

And what do I know about that infinite series?

That infinite series diverges, right?

I can make the partial sums for this as large as I'd like.

As long as I go far enough out in the sequence.

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Well, that means that this series also diverges.

And what does the ratio test say if the limit is equal to one?

When L is equal to one, the series might converge or it might diverge.

The ratio test is just silent in that case.

It doesn't tell us any information. And, and why not?

Well, when L was less than one, then I could pick a value

of epsilon so small that even L plus epsilon was less than one.

And, when L was bigger than one, I could pick a value of

epsilon so small that even big L minus epsilon was bigger than one.

But when L is equal to one, there's no hope for me being

able to choose a small enough epsilon so that I can control the ratio between

subsequent terms, to eventually all be less than

one, or eventually all be bigger than one.

So I'm not going to succeed in comparing my series to a geometric series.

So that's the ratio test.

But finally, a warning. The ratio test tells us to compute this,

the limit of the ratio between subsequent terms.

It's important to emphasize that that limit is not calculating this.