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Welcome the Calculus. I'm Professor Grist.

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

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

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

Â To provide us with two additional tests.

Â For serious convergence these are the root

Â test and the ratio test.

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

Â Indeed from the very beginning of this

Â course, we have understood the geometric series.

Â And when, precisely, it converges.

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

Â convergence tests for more general series. The first of these

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

Â The hypothesis are very minimal. We just need a positive

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

Â following quantity: row, define b, the limit

Â as n goes to infinity of the nth root of

Â a sub n. The test is as follows.

Â The series sum of a sub n converges

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

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

Â if rho equals one? Ha ha.

Â The test fails.

Â It says nothing in that case.

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

Â It

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applies to almost anything. Its ease of use.

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

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

Â limits the usefulness of this test. Rather, when it

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

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

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

Â Power.

Â That looks very difficult with regards to comparisons,

Â or integral tests, or anything of that sort.

Â However, if we apply the root test.

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

Â The

Â [UNKNOWN]

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

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

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

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

Â Well we can evaluate that limit.

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

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

Â That is definitely less than one and that means that

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

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

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

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

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

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

Â We need to compute the limit as n goes to infinity

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

Â minus pth power.

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

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

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

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

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

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

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

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

Â get something that eventually behaves

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

Â is tending to a number that is less than 1,

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

Â The geometric series is well, what if instead of looking

Â at the nth words of the nth term, we looked

Â at the limit then goes to infinity of

Â the ratio between subsequent terms for the geometric series.

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

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

Â We expect similar types of results with this ratio.

Â Indeed, that is the basis of the ratio test.

Â If, as before, we have a positive sequence.

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

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

Â Then, the exact same conclusions hold.

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

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

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

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

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The ratio test should be your first choice.

Â Let's see a few examples.

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

Â cubed over quantity 3n factorial.

Â Now, this could look a little bit confusing.

Â Does this converge, does it diverge? Wow.

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

Â In this case, a sub n equals n factorial cubed

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

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

Â cubed over quantity 3n plus 3 factorial.

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

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

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

Â When we

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

Â The n plus 1 factorial cubed cancels with the n

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

Â Likewise with the 3n factorial and the quantity

Â 3n plus 3 factorial we are left with the

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

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

Â The leading term is cubic in n, and cancels the

Â numerator and the denominator, yielding 127th.

Â 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

Â factorial. Well, you might be able to

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

Â That's certainly what I would guess.

Â Now lets say we needed to show that explicitly.

Â Well we could try the

Â ratio test. Lets see what we get.

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

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

Â Over quantity n plus one factorial.

Â When we compute this limit row, we get n plus

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

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

Â cancellation going on with these factorials.

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

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

Â There's some cancelling that goes on, there.

Â And what

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

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

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

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

Â Those bigger than 1.

Â But it is precisely e, that means this series diverges

Â as we thought. Now sometimes.

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

Â Consider the following complicated looking series.

Â You've got odd numbers, and the numerator and

Â the product of even numbers, and the denominator.

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

Â Well lets try the ratio test.

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

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

Â In the denomiator.

Â When we substitute in n plus 1,

Â we have to up the index on all these terms.

Â But notice the amount of cancellation that happens when we compute

Â the ratio that an plus 1 to an.

Â After all of that cancellation, We're really

Â only left with just a couple of terms.

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

Â 2 times quantity 2n plus 2 in the denominator.

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

Â respectively. That yields a limit of one half.

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

Â You'll learn to what, soon.

Â Sometimes, it's not always clear

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

Â from zero to infinity of pi to 2 3rds

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

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

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

Â The root test tends to work when you have

Â something with an exponent that has an n in it.

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

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

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

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

Â 1, over n factorial. That

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

Â we can use Stirling's formula for the

Â [INAUDIBLE]

Â of the factorial.

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

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

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

Â Zen goes to infinity. Well this one works, and

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

Â That means that not only does this series converge, but

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

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

Â true. N factorial grows much, much more quickly

Â 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,

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

Â Whenever n factorial times something, to the n.

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

Â the x in series form, evaluated at x equals pi

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

Â tests, don't forget to think. Now there

Â are many tests that you have at your disposal.

Â Use them, learn which test tends to be best with

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

Â Keep trying until you unlock the convergence.

Â >> This concludes our principle set of

Â test for determining covariance, divergence of infinite series.

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

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

Â works when.

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

Â In our next lesson we're going to introduce a

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

Â