Welcome to Calculus, I'm Professor Griest. We're about to begin lecture 51, on
convergence tests. >> This lesson will introduce a
collection of tests for determining when an infinite series
converges or diverges. Not all of these tests are useful, in all
circumstances. And it takes some time to learn which test works when. But, with a little bit of practice you'll be discerning convergence and divergence
in no time. >> Our goal in this lesson is to
determine convergence or divergence of a series based on a sequence
as of N. We're going to employ the strategy of
comparison. That is, we'll work with some other
sequence, say, b sub n, that is larger than or smaller than a sub n.
Or instead of another sequence we might use a continuous function that is related
to the original sequence. This is the basis of our first test, the
integral test. A test which has some complicated
hypotheses. First of all, the sequence a sub n, must
be positive and decreasing. Subsequent terms get smaller. Second, we need to take that sequence and
connect the dots, turning it in, to a continuous function, a of x.
Is it itself positive and decreasing and which agrees
with the sequence at every integer point.
If we have both of these, then we can compare the
series, the sum of the a sub n's with the integral if
x goes from 1 to infinity. Of a of x, dx. And these both have the same convergence
or divergence. If one converges, so does the other.
If one diverges, so does the other.
Now, in terms of applicability, this test is not stellar.
It has some strict hypotheses, and in terms of ease of use, this is also not the
simplest test since improper integrals can be tough.
Its overall usefulness is limited. But there are several key examples where
it works well. Why does this test work? Well, we've seen it in use once before, when we determined that the
harmonic series. Diverge. We compared it to the continuous
inter-grand /n 1 over x. We saw, by taking a left riemann sum, that
the series was bounded below by the improper
integral. If instead of a left remonsom we took a right remonsom then we get a bound in the
other direction that the series or at least the
tail, ignoring the first term, bounds from below
the improper integral. Having this bound on both sides means that
the series and the integral have the same convergence or
divergence properties. If we look at a few examples, we'll see
better how to apply this. Consider the absolutely fundamental
series, sum 1 to infinity, of 1 over n to the p, where p is
a constant. This should remind you of the p integrals
from when we did improper integrals.
In fact these are called the p series. And the right way to understand their
convergence or divergence is to compare to the
integral. As x goes from one to infinity of one over
x to the p. Dx. Now we know, that these p integrals
converge when p is strictly greater than 1.
And diverge when p is less than or equal to 1.
Therefore, the same holds for the p series.
This is a set of examples. You must memorize, a related example that
is shown in the same manner.
Is the sum of 1 over n times log of n to the pth power.
If we continue this to the function.
1 over x times log of x to the p. Then we can integrate that with respect to
dx. And obtain an integral using a little u
substitution to get log of x to the 1 minus p over 1 minus p. Evaluating those, we see.
Then again, we get convergence when p is bigger than 1 and
divergence when p is less than or equal to 1. Thus, the same convergence and divergence
holds for this series. And in both these cases, we have a range
of convergence for p. Where it's strictly greater than one.
Our next test is a comparison test between two sequences.
Our hypothesis are relatively simple. We have two sequences, a sub n and b sub
n. Both are positive, and the b sub n's are
bigger than the a sub n's. If this is true, then what can you say about the series?
Well, if b sub n is bigger than or equal to a sub n, then the sum of the b
sub n's is bigger than or equal to the sum of a sub n's.
That much is clear. But what can we say about convergence or
divergence? Well, because all the terms are positive,
if the larger of the two series converges, then the smaller one
must converge as well. Now, the converse of this is not.
Necessarily, so but the [UNKNOWN] positive is if the smaller series
diverges, then so does the larger. Now in terms of applicability oh, this one
doesn't have so many hypothesis. This is the integral test. But in terms of ease of use and
usefulness, well it's not perfect. In part, because it's easy to get these
two results confused, and it's sometimes difficult to
pick the appropriate b sub n, or a sub n, or to know which is
which. Let's see this in the context of some
examples. If we look at the sum, as n goes from 0 to
infinity, of 1 over quantity, 4 plus negative 1 to the n,
to the n. That's 1 plus a third plus 1 over 5
squared, plus 1 over 3 cubed, plus 1 over 5 to the fourth
and etcetera. That's a funny looking series but it's not
that hard to figure out because if we called this terms
the a sub n's, they are definitely bounded above by a sequence b
sub n, given by one of our 3 to the n.
Summing that up gives us a geometric series, which definitely converges. Therefore, the a series converges as well.
Now, one of the problems is no one told us which is a, which is b,
and what we should use. We could.
Have bounded the a sub n by a p series.
Where p in this case is equal to 2. That p series converges and so the
comparison test would tell us that the original series
converged as well. On the other hand, a perfectly good upper
bound is given by 1 over n.
However, this p series diverges and it doesn't tell us anything about whether the a series converges or
not. This is the subtlety of using the
comparison test. P series tend to be extremely helpful when
using the comparison test. Consider the sum from 0 to infinity of
cosine squared n over 1 plus the square root of n cubed.
Let's let that be our a sub n. We can bound the numerator. From above by one since cosine squared is
always between 0 and 1. What can we do with that denominator?
Well, the denominator is strictly larger than n to the 3 halves.
Therefore, we can bound this a sequence by a b sequence of the
form, 1 over n to the 3 halves.
Since we know something about the p series, for p equals 3 halves, we know
that the larger series converges.
Therefore, the smaller a series converges as well.
Now let's see an example where it runs in the other direction.
Consider the sum of our n of 1 plus square root of n.
Over square root of 1 plus n cubed. Now if we let that term be our a sub n,
then we can bound the numerator by, let's say, 2 square root of n.
And the denominator. By square root of n cubed.
Now, that's going to give us a b sub n that is a p series with p equals 1. That is not going to help us.
That series diverges. It doesn't tell us anything about the a
series. So, what are we going to do? Well, we implicitly assumed that our given
was the a. But there's no reason why that has to be
the case. Well let's let this be the b terms. And we can choose an a term that bounds
from below. Taking a numerator of the square root of
n, and a denominator of let's say 2n cubed.
Then we get a lower bound for the b terms, since the a terms give us a p
series with p equals 1.
The a series diverges and thus does the b series diverge as well. Now another example leads us to some
complexity if we inlet a sub n b to n cube plus 3 n
minus 8 over n to the fifth minus 5 n cubed
minus n squared plus 2. And what are we going to choose, when we
try to do the comparison test? Well, we need to bound the numerator from
above. Let's say 2 n cubed, no that's not going
to work, because we've got the 3 n. Let's say 3 n cubed.
That will do the job. The denominator, well we can bound from below by 2n to the
fifth? No. N to the fourth? No, that's not going to be good, that
gives us a harmonic series for the v sub n, that
diverges. Ugh, this gets very complicated. Maybe, the comparison test is the wrong
one. Instead, let's try an integral test.
Who wants to integrate 2x cubed plus 3x minus 8 over x to the 5th minus 5x
cubed minus x squared plus 2. Hello, class?
Where did everyone go? Well, what are we going to do in this
case? Asymptotic analysis is going to save us as
it has done so many times before. Taking this, a of x we see that the
leading order term is 2 over x squared. Everything else is in big O of 1 over x to
the 4th. This x goes to infinity. Well the same thing holds for the discrete
setting with our sequence a sub n.
Since the leading order term, is a p series with p plus 2, we know that that
converges. Therefore, we know that this series
converges because we know the leading order term. That's the idea behind our last test in
this lesson, the limit test. The limit test says, if you have 2
positive sequences that have the same asymptotics, the same
meeting order term/g. Specifically, if their limit, of the
quotient is n goes to infinity, is some number that is strictly between 0
and infinity, then these two series have the same convergence or
divergence behavior. This is a wonderful test, very easy to
apply and to use. And it's overall usefulness is very high
as well. Let's see some examples where we compute
the leading order terms. Just as we did with improper integrals. Consider the sum over n of log squared of
quantity 1 plus 1 over n squared. That looks very complicated to figure out, I don't want to try an integral test with
that, I don't know what to compare it to. But I do know the Taylor series for log of
1 plus something. But that something is very small. I know that the leading order term for
this seqeunce is 1 over n squared quantity squared.
That is 1 over n to the 4th. And that's a p series, with p equals 4, it converges.
Thus, does my original series converge? Here's another crazy looking example,
we'll put a polynomial on top, and something with some square roots
in it down below. All we need to do, is compute the leading
order term for the numerator, it's an n squared with a 3 in front but
forget about the constant. In the denominator, the leading order
term, is n times n to the 5 3rds. When you get the exponents right, you see
that the leading order term is 1 over m to the 2
3rds. That's a p series with p equals 2 3rds. [SOUND] forget it. That diverges and so does the original
series. Lastly, consider n times pi to the n over
the hyperbolic cosine of n. Well the leading order term in hyperbolic
cosine is exponential in n. Therefore, with a little bit of factoring,
we see that the leading-order term is n, times pi over e, to the n.
This definitely diverges. Always look to compute the leading-order
term. >> All of the tests that we've
considered in this lesson are based on some form of comparison and
ultimately reduced to finding the [INAUDIBLE] or leading number term in the sum. Isn't it a good thing that we're familiar
with [INAUDIBLE] from earlier in the course. In our next lesson we'll look at a
different collection of tests. Ones that are based on the geometric
series.