### Video Transcript

In this lesson, weβll learn how to
determine whether a series is convergent or divergent by comparing it to a series of
known convergence using the limit comparison test. Weβll look at a number of examples
of how we can apply this test. But first, letβs look at how the
test works.

Remember that the series is the sum
of the terms of an infinite sequence of numbers. If we have two series where the
terms are π π and π π, where π π and π π are greater than zero for all
π. The limit comparison test tells us
that if the limit as π tends to β of π π over π π is the constant πΆ, where πΆ
is positive and finite, then either both series converge or both series diverge. Note that it doesnβt matter which
of the two series is the numerator and which is the denominator in the limit. What this test says that as π
tends to β, the terms of the two series π π and π π behave in a similar way. So we can expect the series
themselves to exhibit similar behaviour to each other. Note that the limit is often
written as, the limit as π tends to β of π π multiplied by one over π π which
is equal to πΆ.

The limit comparison test is a
variant on the direct comparison test, which we can use when we canβt apply the
direct comparison test. In practice, how we use the limit
comparison test is if weβre asked to determine the convergence of a particular
series, say π π, summed from π equals one to β. We bring in the series π π,
summed from one to β, which we know the convergence of. We use the terms of the two series
to find the limit πΆ, which allows us to state the convergence or otherwise of our
series, π π from π is one to β. Letβs apply this in an example.

Use the Limit comparison test to
determine whether the series the sum from π is one to β of two over π plus six is
convergent or divergent.

Weβve been given the series whose
terms weβll call π π, which is the sum from π is one to β of two over π plus
six. And weβre asked to determine the
convergence of this series using the limit comparison test. The limit comparison test tells us
that for series with terms π π and π π, where π π and π π are greater than
zero. If the limit as π tends to β of π
π over π π is a constant πΆ, where πΆ is greater than zero and finite, then
either both series converge or both diverge. Weβve been given the series where
the terms are two divided by π plus six. And since the two is a constant, we
can take this outside the series so that we have two times the sum from π is one to
β of one over π plus six.

To determine whether this series
converges or diverges, we need to find another series with easily determined or
known convergence, such that the limit comparison test applies. Notice that the limit comparison
test is concerned with the terms of the series, not the series themselves. And it looks at the behaviour of
the ratio of these terms as π tends to β. So letβs look at the terms in our
original series as π tends to β.

Since weβve taken the two outside,
our term is now one over π plus six. And as π tends to β, this term is
dominated by the π in the denominator. The six becomes superfluous as π
gets very large. And our term has the same behaviour
as one over π. We can therefore use the sum from
π is one to β of one over π as our second series. You might recognize this as the
harmonic series. This is related to the wavelengths
of the overtones of a vibrating string. So now we have our two series. We can think about applying the
limit comparison test. But first, we need to check that
the terms of each series are positive.

If our term is π π is equal to
one over π and π is positive from one to β, then one over π is always
positive. Similarly, if our term is one over
π plus six, since π plus six is always positive, one over π plus six is positive
too. So both terms are positive, and we
can use the limit comparison test.

Our first step is to find the limit
as π tends to β of π π over π π, with π π is one over π and π π is one
over π plus six. Thatβs the limit as π tends to β
of one over π divided by one over π plus six. Thatβs equal to the limit as π
tends to β of one over π multiplied by π plus six over one. Which is the limit as π tends to β
of π plus six divided by π. And thatβs equal to the limit as π
tends to β of one plus six divided by π. We know that as π tends to β, six
divided by π tends to zero. So that our limit is actually equal
to one, which is a positive constant. So weβve found our constant πΆ,
which is greater than zero. Then thatβs the limit as π tends
to β of π π over π π.

So all we need to know now is the
convergence of our second series, the harmonic series. You probably know that this series
diverges, but we can use the integral test to show this. The integral test tells us that if
π is continuous positive and decreasing on the interval one to β and we let π π
equal to π of π. Then if the improper integral from
one to β of π of π₯ dπ₯ converges, so does the sum from π is one to β of π
π. Alternatively, if the improper
integral from one to β of π of π₯ with respect to π₯ diverges. Then so does the sum from π is one
to β over π π. Before we can use this test, we
need to check that our terms are positive and decreasing.

One over π is positive for all
positive π, so that π π is greater than zero. We know also that for positive π,
π is less than π plus one. And taking the reciprocal, we
reversed the inequality. So that one over π is greater than
one over π plus one. So that π π is greater than π π
plus one. So that as π increases, π π
decreases. So then, we have that the sum from
π is one to β of one over π and the integral between one and β of one over π₯ with
respect to π₯, either both converge or both diverge.

If we work out this integral, the
integral of one over π₯ is the natural log of π₯. And we evaluate this between one
and β. And thatβs β minus zero, which is
β. Our integral therefore
diverges. So our sum from π is one to β of
one over π also diverges. And hence, by the limit comparison
test, our first series, which is the sum from π is one to β of two over π plus six
also diverges.

So we found a second series to use
in the limit comparison test. And we found the limit as π tends
to β of the ratio of the terms of the two series is a positive constant. We found that the second series
diverges. And therefore, so does our original
series.

Letβs try another example.

Using the limit comparison test,
determine whether the series the sum from π is one to β of one over four to the
power π plus six is convergent or divergent.

Weβve been given a series the sum
from π is one to β of one over four to the π plus six. The terms of our series are
therefore one over four to the π plus six. And weβre asked to determine the
convergence of this series using the limit comparison test. This test says that if we have two
series with terms π π and π π, which are positive, and the limit as π tends to
β of π π over π π is a constant πΆ, which is greater than zero and less than
β. Then either both series converge or
both series diverge.

We have a series with terms one
over four to the π plus six. In order to use the limit
comparison test, we need another series to compare in our ratio of terms. Weβd like to find another series
where the terms will make it easier to show that one term divided by the other as π
tends to β is constant. We also need to know whether this
second series converges or diverges. To find the terms of this second
series, what we can do is look at the behaviour of our terms as π tends to β. As π tends to β now in term one
over four to the π plus six, the six becomes superfluous, and our term turns to one
over four to the power π.

This could help us since this is a
term of a geometric series. So if we let the terms of our first
series equal to π π, thatβs one over four to the π plus six is equal to π π,
and the terms of our second series, one over four to the π equal π π. We can then find the limit as π
tends to β of π π over π π within our limit comparison test. Note that it doesnβt matter which
one is π π and which one is π π, whichever makes the calculation easiest. So weβre going to find the limit as
π tends to β of π π over π π, where π π is one over four to the π and π π
is one over four to the π plus six.

So we have the limit as π tends to
β of the one over four to the π divided by one over four to the π plus six. And thatβs equal to the limit as π
tends to β of the one over four to the π times four to the π plus six divided by
one. Thatβs the limit as π tends to β
of four to the π plus six divided by four to the π. Which is the limit as π tends to β
of one plus six, divided by four to the π. Since one is a constant, thatβs one
plus the limit as π tends to β of six divided by four to the power π. And six is a constant. We could take this outside, so that
gives us one plus six times the limit as π tends to β of one over four to the
π. And as π tends to β, one over four
to the power π tends to zero.

Our limit is therefore one which is
a constant greater than zero. So we found a second series where
the ratio of the terms of the two series as the π tends to β is a constant. Now, we need to find the
convergence of that second series. Because by the limit comparison
test, the convergence of the second series determines the convergence or divergence
of the first series. Our second series is the sum from
π is one to β of one over four to the power π. And we know that this is a
geometric series. A geometric series is a series with
the sum from π is zero to β of π times π to the power π. And thatβs π times π to the zero
plus π times π to the one plus π times π squared, et cetera. If the absolute value of π is less
than one, then the series sums to π over one minus π. If the absolute value of π is
greater than or equal to one, then the series diverges.

In our series, π is equal to one,
and π is equal to one over four. There is a difference, however. For the defined geometric series,
π begins with the value of zero. Whereas in our series, π starts at
one. Given that π, in our case, is less
than one, when we come to work out the sum of our series, weβll need to subtract
this first term. So letβs work out the sum of our
series, which is π over one minus π. The first term in the defined
series π times π to the power zero, in our case, will be one times one over four
to the power zero, And thatβs equal to one. So weβll need to subtract this from
the sum of our series.

That means we subtracted from π
over one minus π where, in our case, π is one and π is one over four. The sum of our series is therefore
the sum from π is one to β of one over four to the power π, which is equal to one
over one minus one over four minus one. Where this negative one corresponds
to π times π to the power zero. The sum of our second series is
therefore four over three minus one which is one over three. Our second series, the sum from π
is one to β of one over four to the power π, converges to the sum one over
three. By the limit comparison test, if
one of our series converges, then so does the other. This means that our original
series, the sum from π is one to β of one over four to the power π plus six, also
converges.

Letβs look at one more example of
the limit comparison test for the convergence or divergence of a series.

Use the Limit comparison test to
determine whether the series the sum from π is one to β of three π squared plus
two π plus one divided by π to the power three plus one is convergent or
divergent.

Weβve been given a series where the
terms are π π equal to three π squared plus two π plus one divided by π to the
power three plus one. Weβre asked to use the limit
comparison test to determine whether the series converges or diverges. The limit comparison test tells us
that if we have two series where the terms π π and π π are positive and the
limit as π tends to β of π π divided by π π is a constant πΆ such that πΆ is
greater than zero and less than β, so itβs finite. Then either both the series
converge or both series diverge. In order to use this test to
determine whether the given series is convergent or divergent. Our plan of action is to try and
find a second series such that the ratio of the terms π π over π π converges to
a positive number.

We also need to know or be able to
find out whether the second series converges or diverges. To find the second series, we can
look at the behaviour of these terms of our first series as π tends to β. However, letβs make sure our terms
are positive before we start. Since π is positive, both the
numerator and the denominator are greater than zero. So in fact, our terms are
positive. We can therefore use the limit
comparison test. For large π in this expression,
the terms with the highest powers of π dominate. This means in the numerator, three
π squared dominates. While in the denominator, π to the
power three dominates. So as π tends to β, the terms of
our series tend to three π squared divided by π to the power of three. And thatβs equal to three over
π.

We can use this as the term in our
second series for the limit comparison test. Our second series is therefore the
sum from π is one to β of three divided by π. Now, with π π equal to three π
squared plus two π plus one divided by π to the power of three plus one and π π
equal to three divided by π, we can find the limit within our limit comparison
test. Thatβs the limit as π tends to β
of π π over π π. Thatβs equal to the limit as π
tends to β of three π squared plus two π plus one divided by π to the power of
three plus one all divided by three divided by π. Thatβs equal to the limit as π
tends to β of three π squared plus two π plus one divided by π to the power of
three plus one multiplied by π divided by three.

Thatβs equal to the limit as π
tends to β of three π to the power of three plus two π squared plus π all divided
by three π to the power of three plus three. If we divide both the numerator and
the denominator by three π to the power of three. That gives us the limit as π tends
to β of one plus two π squared over three π to the power of three plus π over
three π to the power of three all divided by one plus three over three π to the
power of three. And cancelling down that gives us
the limit as π tends to β of one plus two over three π plus one over three π
squared all divided by one plus one over π to the power of three.

Two divided by three π tends to
zero as π tends to β. One over three π squared tends to
zero as π tends to β. And one over π to the power of
three tends to zero as π tends to β. This means that the limit as π
tends to β of π π over π π is equal to one. This is a positive constant, so
satisfies our limit comparison test. So now if we can determine the
convergence of our second series, this will establish the convergence or divergence
of our first series.

Our second series is the sum from
π is one to β of three divided by π. And since three is a constant
multiply, we can take this outside. This gives us three times the sum
from π is one to β of one over π. The sum from π is one to β of one
over π exactly the harmonic series. And we know that this series
diverges. And this means by the limit
comparison test that our first series diverges also. The series which is the sum from π
is one to β of three π squared plus two π plus one divided by π to the power of
three plus one diverges.

Weβve looked at some examples of
the limit comparison test. Now, letβs summarize with some key
points. The limit comparison test tells us
that if we have two series with positive terms π π and π π and the limit as π
tends to β of the ratio of the two terms is a constant πΆ, where πΆ is positive and
finite, then either both series converge or both series diverge. To use the limit comparison test to
determine the convergence or divergence of a given series.

We try to find a second series of
known convergence or divergence and whose terms all satisfy the condition that the
limit as π tends to β of the ratio of the terms is equal to a constant. We can often find the second series
by looking at the behaviour of the terms of the given series as π tends to β. Finally, we note that when we
calculate the limit as π tends to β of π π over π π, it doesnβt matter which we
choose as π π or π π.