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 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 [UNKNOWN] 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. 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. 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 [INAUDIBLE] 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. 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.