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Welcome the Calculus. I'm Professor Grist.

We're about to begin Lecture 52, on Convergence Tests Part two.

>> One of the first series that we saw in this course was the geometric series.

In this lesson, we're going to leverage our understanding of the geometric series.

To provide us with two additional tests.

For serious convergence these are the root

test and the ratio test.

>> There is one series whose convergence and divergence we understand well.

Indeed from the very beginning of this

course, we have understood the geometric series.

And when, precisely, it converges.

We're going to use that as the basis for two

convergence tests for more general series. The first of these

tests is called the root test. It goes as follows:

The hypothesis are very minimal. We just need a positive

sequence, a sub n. From that, we compute the

following quantity: row, define b, the limit

as n goes to infinity of the nth root of

a sub n. The test is as follows.

The series sum of a sub n converges

if row is less than one. If row is greater than

one, then the series diverges. Now you might wonder, what happens

if rho equals one? Ha ha.

The test fails.

It says nothing in that case.

Now as far as applicability, why, this is a very general test.

It

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applies to almost anything. Its ease of use.

Well, it depends. Computing the nth root and taking a limit

as n goes to infinity, that can be a little difficult, and that

limits the usefulness of this test. Rather, when it

does work, it does what other tests usually cannot do.

Let's see some examples. Consider the sum n goes from 1 to infinity

of quantity log of 1 plus cosine 1 over n, all raised to the one half n.

Power.

That looks very difficult with regards to comparisons,

or integral tests, or anything of that sort.

However, if we apply the root test.

Then we compute row, the limit as the end goes to infinity.

The

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root of a sub n. And since a sub n involves an

exponent with an n in it, this nth root cancels that.

And we're left with the limit as n goes to infinity.

The square root of log of quantity 1 plus cosine 1 over n.

Well we can evaluate that limit.

Zen goes to infinity cosine of 1 over n is tending towards 1.

Therefore row is equal to the square root of the log of 2.

That is definitely less than one and that means that

this series converges, was that obvious to you from the beginning?

Wasn't obvious to me. Well, here's another example.

Let's look at the p sereis, where a sub n

is one over n to the p for some constant p.

In this case, when we compute row, what happens?

Well, we see some of the limitations of the root test.

We need to compute the limit as n goes to infinity

of the nth root of n, and then raise that to the

minus pth power.

Well, that limit of the nth root of n is equal to 1.

And one to anything is one. This test fails for the p series.

Of course, this must be the case. Because depending on the value of p.

The p series either converges or diverges. In this case, the root

test doesn't tell you. Now, the reason why the

root test works is through a comparison to the geometric series.

In the geometric series, a sub n is really x to the

n, and we see that in taking the nth root. If we

get something that eventually behaves

like a geometric series, that is where the nth root of a7

is tending to a number that is less than 1,

then we get convergence. But there's another way that we could use.

The geometric series is well, what if instead of looking

at the nth words of the nth term, we looked

at the limit then goes to infinity of

the ratio between subsequent terms for the geometric series.

A sub n plus 1, divided by a sub n, is really x

to the n plus 1, divided by x to the n, that is x.

We expect similar types of results with this ratio.

Indeed, that is the basis of the ratio test.

If, as before, we have a positive sequence.

And we compute row, this time, to be the limit as n goes

to infinity of as of n plus 1 over as of n.

Then, the exact same conclusions hold.

When this quantity is less than 1, the series converges.

And this quantity is greater than 1. The series diverges.

Now, this is a test. This test tends to be extremely easy to

use and extremely useful. So much so, that for most series.

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The ratio test should be your first choice.

Let's see a few examples.

Consider the sum and goes from 0 to infinity, of, n factorial

cubed over quantity 3n factorial.

Now, this could look a little bit confusing.

Does this converge, does it diverge? Wow.

Let's see what the ratio test has to say.

In this case, a sub n equals n factorial cubed

over quantity 3n factorial. A sub n plus 1, well we have to be careful

when we're doing factorials. This is n plus 1 factorial quantity

cubed over quantity 3n plus 3 factorial.

Now, when we compute rho. That is, the limit of the ratio.

Remember, it's a sub n plus 1. On top, a sub n, below, we can

evaluate this by multiplying a sub n plus 1, times the reciprocal of a sub n.

When we

do so, because of these factorials, there's a great deal of cancellation.

The n plus 1 factorial cubed cancels with the n

factorial cubed, leaving just an n plus 1 quantity cubed.

Likewise with the 3n factorial and the quantity

3n plus 3 factorial we are left with the

limit, as n goes to infinity, and n plus one quantity cubed, or at 3n

plus 3, times 3n plus 2, times 3n plus 1.

The leading term is cubic in n, and cancels the

numerator and the denominator, yielding 127th.

That is much less than one, and we have a convergent series.

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Here's another example. Consider the sum of n to the n over n

factorial. Well, you might be able to

argue, based on what you know about n to the n that this diverges.

That's certainly what I would guess.

Now lets say we needed to show that explicitly.

Well we could try the

ratio test. Lets see what we get.

A sub n is a to the n over n factorial.

A sub n plus one is n plus 1 quantity to the n plus 1.

Over quantity n plus one factorial.

When we compute this limit row, we get n plus

one to the n plus one over n plus one factorial times n factorial

over n to the n. As before, there's quite a bit of

cancellation going on with these factorials.

And we're left with, the limit, as n goes to infinity

of n plus 1 to the n plus 1 over n plus 1, times 1 over n to the n.

There's some cancelling that goes on, there.

And what

do we get? we get, after a little bit of rewriting a

limit that we've seen before. The limit is n goes to infinity of

quantity one plus one over n all to the f, that is e.

That means the ratio of the subsequent terms tends to a number.

Those bigger than 1.

But it is precisely e, that means this series diverges

as we thought. Now sometimes.

It's not clear exactly which test ought to be used.

Consider the following complicated looking series.

You've got odd numbers, and the numerator and

the product of even numbers, and the denominator.

And one over two to the n. What do we do?

Well lets try the ratio test.

In this case a sub n is the product of odd numbers up to, 2 n minus 1 divided

by the product of even numbers up two n with an extra 1 over 2 to the n.

In the denomiator.

When we substitute in n plus 1,

we have to up the index on all these terms.

But notice the amount of cancellation that happens when we compute

the ratio that an plus 1 to an.

After all of that cancellation, We're really

only left with just a couple of terms.

That would be 2n plus 1 in the numerator, and

2 times quantity 2n plus 2 in the denominator.

The leading order terms in numerator and denominator are 2n and 4n

respectively. That yields a limit of one half.

Therefore this series converges. It converges to something very nice.

You'll learn to what, soon.

Sometimes, it's not always clear

which test is going to be best. Consider the sum as and goes

from zero to infinity of pi to 2 3rds

n divided by e to the n times n factorial. What

is this going to be? It seems like it ought to converge.

Hm, let's see. We could try the root test.

The root test tends to work when you have

something with an exponent that has an n in it.

Let's see what happens if we take the limit as n goes to infinity of the nth

root of pi to the two thirds n divided by e of the n times n factorial.

Well, that pi term and the e term work out nicely but

we're left with pie to the two thirds over e, times the nth root of

1, over n factorial. That

is maybe not so easy to do, if you want to get fancy then

we can use Stirling's formula for the

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of the factorial.

Wow I don't know I'm not really in the mood for that.

I'm not quite sure how to valuate that limit so let's try the ratio test

instead we take ace of n plus one multiplied by ace of n and take the limit.

Zen goes to infinity. Well this one works, and

it works well. We see that row as a limit, is equal to 0.

That means that not only does this series converge, but

the terms in the series get small very, very quickly.

Now, if we've thought about that, we would have said, well, of course that's

true. N factorial grows much, much more quickly

than pi to the 2 3rds n. So, certainly this must converge.

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On the other hand, if we had been thinking from the beginning,

we might have noticed that this series is actually of the form some.

Whenever n factorial times something, to the n.

That means we're really looking at our old friend, e to

the x in series form, evaluated at x equals pi

to the 2 3rds over e, once you start doing convergence

tests, don't forget to think. Now there

are many tests that you have at your disposal.

Use them, learn which test tends to be best with

which type of series. And then, like a collection of keys,

Keep trying until you unlock the convergence.

>> This concludes our principle set of

test for determining covariance, divergence of infinite series.

At this point it would be a good idea to do a

lot of homework problems so that you gain an intuition of which test

works when.

Just like with integration, not every method is useful in all circumstances.

In our next lesson we're going to introduce a

new class infinite series, and refine our notions of convergence.