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

N, log N.

[SOUND]

The harmonic series diverges, but the sum of the reciprocals

of the squares converges. P series with P equals 2 converges.

Alright. So this series converges.

The harmonic series diverges, so this series diverges.

But what's something in between, right?

Where's the boundary between convergence and divergence?

Well here's an example of a series that sits in between.

Its the sum N goes from 4 to infinity of 1 over N log N.

Now, 1 over N log N is less than 1 over

N, because N times log N is bigger than N, right?

A bigger denominator makes the fraction smaller.

And because N squared is bigger than N log N, 1

over N squared is less than 1 over N log N.

So, this thing is bigger

than something that converges, smaller than something that

diverges, so it's, it's really a question, right?

Does this series converge or diverge?

We can use Cauchy Condensation.

So let's analyze this series using Cauchy Condensation.

So I'll write down the condensed series.

It's the sum N goes from 2 to infinity of 2 to the N times the 2 to the Nth term.

So 2 to the Nth term

mean I put a 2 to the N times. Log 2 to the N in the denominator.

Okay, so this is the condensed series, this is the original series.

Cauchy Condensation tells me that these two series share the same fate.

They either both converge or both diverge.

So it's enough to figure out what's going on with this series.

Oh and look!

I've got a 2 to the N in the numerator and a 2 to

the N in the denominator, so we cancel those and we can rewrite this as

the sum N goes from 2 to infinity, which is 1 over log 2 to the N.

Now I could use property of logs to simplify this.

This is the sum N goes from 2 to infinity of 1 over N times log 2.

That's just because log 2 to the N is N times log 2.

And then, well look at this.

This is

1 over log 2 times the sum N goes from 2 infinity of 1 over N.

Oh, but this is bad or really great news depending on your attitude.

This thing here, is the tail of a harmonic series.

So that means that the condensed series diverges.

And because the condensed series diverges, this original series must also diverge.

If you like integrals, you can use an integral test.

Okay.

So let's do this with the integral test.

So I could look at this integral, the integral from

4 to infinity of 1 over X times log X DX.

That's a suitable function to consider.

By definition, this integral from 4 to infinity just means the limit as N

approaches infinity of the integral from 4 to big N of 1 over

X times log X DX.

But now, how do I evaluate that definite integral?

Well, I happen to know this.

The derivative of log of log of X is what?

Well by the chain rule, this is the

derivative log, which is 1 over, evaluate the

inside, log of X, times the derivative of the inside function, which is 1 over X.

Well, here I've got 1 over

X times log X. Here, I've got some other function.

Here I've got an anti-derivative for this integrand.

So, that's enough for me to be able to

calculate this definite integral using the fundamental theorem of calculus.

So this is the limit as N approaches infinity of what is this?

This is telling me an anti-derivative, so log log X evaluated

at 4 and N.

And then I could plug in N and plug in 4 and take the difference.

This is the limit N goes to infinity of log log N minus log log 4.

What is this limit?

Well, it's growing very slowly but it is, in fact, running off to infinity, right.

By choosing N big enough, I can make log N as large as I like,

and that means I can also make log log N as large as I like.

So this limit is in fact, infinity.

That means that this integral diverges.

That means that the original series diverges as well.

So that diverges, but maybe if we mess around

with it a bit we can make it converge.

So our original question was whether or not this series

converged or diverged.

And now we've seen, both by using condensation

and the integral test, that this series diverges.

We're going to go back to our original story, right?

We were thinking of this series as somehow sitting in between these two series.

But now where's the boundary between convergence and divergence, right?

This series converges, these two series diverge.

So let me try to fit another

series in between these two series. Here's an example.

Its the sum N goes from 4 to infinity of 1

over N times log N, and that log N is squared.

Does that series converge or diverge?

We can use condensation. So lets write down the condensed series.

In this case, this is the sum N goes from 2 to infinity of 2

to the N times the 2 to the Nth term.

So, 2 to the N times log 2 to the N squared.

Now, how do I evaluate this series, right?

If I can determine the convergence of this

series, I've determined the convergence of the original series.

Good news.

This 2 to the N and this 2 to the N cancel.

So now I'm left with the sum N goes from 2 to

infinity of 1 over log 2 to

the N squared. What else can I do there?

Oh, I can use properties of logs again. So this is the sum N goes

from 2 to infinity of 1 over N times log 2 squared.

I could factor out the 1 over log 2 squared.

So this is 1 over log 2 squared times the sum N goes from

2 to infinity of 1 over N squared. this is really great.

Right?

because what do I know? This is a P series, where P equals 2.

This series converges.

Well, that means that this condense series converges.

That means the original series converges. So, we've got a convergence

series and a divergence series. It's worth comparing these two series.

In this series, which we were asking

about, we now know converges by Cauchy condensation.

And we've already seen a little while ago that this series diverges.

So again, we could try to play the same game, there's

just something that we try to fit in between these two.

Well, here's our question.

Does this series, which sort of fits

in between these two series, does this series converge or diverge?

This is the series N goes from 4 to infinity

over 1 over N times log N times log log N.

Well, I'll leave it to you to analyze this series.

[SOUND]

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