### Video Transcript

Comparison Test for Series

In this video, we will learn how to
determine whether a series is convergent or divergent by comparing it to a series of
known convergence using the comparison test. Weβll be looking at a variety of
examples of how we can do this. First, letβs cover how the
comparison test for series works. When using the comparison test,
there are two series which we are concerned with. We can call these series the sum
from π equals one to β of π π and the sum from π equals one to β
of π π.

Now, in order to use the comparison
test, we require both π π and π π to be greater than or equal to zero for all
π. We are also required to know
whether the sum from π equals one to β of π π converges or diverges. Now, the comparison test tells us
that if the sum from π equals one to β of π π converges, and π π is less
than or equal to π π for all π, then the sum from π equals one to β of
ππ also converges. Conversely, if the sum from π
equals one to β of π π diverges, and π π is greater than or equal to π
π for all π, then the sum from π equals one to β of π π also
diverges.

Now, itβs quite intuitive to see
how the comparison test works. Since in the first part, weβre
saying that if all our terms are smaller than or equal to the terms of a series
which we know to be convergent, then our series also converges. And in the second part, weβre
saying that if our terms are larger than or equal to the terms of a series which we
know that diverges, then our series also diverges. Now, using the comparison test
requires us to know a few convergent and divergent series.

Since we need to know some series
which are convergent or divergent, so thatβs the series π π, which we can use to
compare to the series which weβre trying to find the convergence of. So, thatβs the series of π π. Now, one type of series which is
often quite useful when using the comparison test is a π series. A π series has the form of the sum
from π equals one to β of one over π to the π. And the π series test will tell us
that this series will diverge if π is less than or equal to one. And it will converge if π is
greater than one. Letβs now move on and look at an
example of how we can use the comparison test.

Use the comparison test to
determine whether the series which is the sum from π equals one to β of the
natural logarithm of π over π is convergent or divergent.

We can start answering this
question by recalling what the comparison test tells us. The comparison test tells us that
for the series which is the sum from π equals one to β of π π and the sum
from π equals one to β of π π, where π π and π π are both greater than
or equal to zero for all π. That if the sum from π equals one
to β of π π is convergent, and π π is less than or equal to π π for all
π, then the sum from π equals one to β of π π is also a convergent. And if the sum from π equals one
to β of π π is divergent, and π π is greater than or equal to π π for
all π, then the sum from π equals one to β of π π is also divergent.

Now, the series weβve been given in
the question is the sum from π equals one to β of the natural algorithm of
π over π. Therefore, we can let π π be
equal to the natural logarithm of π over π. Now, we need to find another series
to compare this to. Looking at π π, we can see that
it looks similar to a π series. Now, a π series is a series of the
form of the sum from π equals one to β of one over π to the π. And this π series will diverge if
π is less than or equal to one and converge if π is greater than one.

Now, the π series, which our
series looks most similar to, is the sum from π equals one to β of one over
π. For this series, the value of π is
one. And therefore, the series will
diverge. Before we can attempt to use the
comparison test, we first need to check whether π π and π π are greater than or
equal to zero for all π. Now, the values of π which weβll
be using go from one to β. Therefore, π is greater than or
equal to zero.

Now, the natural logarithm of π is
an increasing function. And when π is equal to one, the
natural logarithm of π is equal to zero. Therefore, we can say that for all
our values of π, the natural logarithm of π will be greater than or equal to
zero. Since these two things are both
greater than or equal to zero, this means that both π π and π π will be greater
than or equal to zero too.

Now, weβve already said that the
sum from π equals one to β of one over π is divergent. Therefore, we need to try and
satisfy the second part of the comparison test. Which means that we need to show
that π π is greater than or equal to π π. In other words, thatβs showing that
the natural logarithm of π over π is greater than or equal to one over π. Since π is always greater than or
equal to one, we can multiply both sides by π. Therefore, we have that the natural
logarithm of π is greater than or equal to one. And we need to try and prove this
for all π.

Now, we can immediately see that
this cannot be the case since the natural logarithm of one is equal to zero, which
is less than one. However, this doesnβt mean that we
cannot use the comparison test. Letβs consider the first few values
of the natural logarithm of π. We have that the natural logarithm
of two is equal to 0.693 and so on. And the natural logarithm of three
is equal to 1.098 and so on.

Therefore, the natural logarithm of
three is in fact greater than or equal to one. And as we said earlier, we know
that the natural logarithm of π is an increasing function. And so, our statement that the
natural logarithm of π is greater than or equal to one holds true for values of π
which are greater than or equal to three.

Now, we can rewrite our series
which we were given in the question. We can take the first two terms out
of the series. And we have that the sum from π
equals one to β of the natural logarithm of π over π is equal to the
natural logarithm of one over one plus the natural logarithm of two over two plus
the sum from π equals three to β of the natural logarithm of π over π. Now, we can compare this series
with the π series of the sum from π equals three to β of one over π.

We have the same values of π π
and π π from before. And again, weβve of course
satisfied the condition that π π and π π are both greater than or equal to zero
for all π. And we have that the sum from π
equals one to β of π π is divergent. And we also now have that π π is
greater than or equal to π π for all π. And so, from this, we can say that
the sum from π equals three to β of the natural logarithm of π over π is
divergent.

We can now relate this back to the
original sum we were given in the question since its equal to two constant terms
plus the series weβve just shown to be divergent. Since itβs equal to the sum of two
constants and a divergent series, we have found that the sum from π equals one to
β of the natural logarithm of π over π is divergent.

Letβs now look at a second example
of how we can use the comparison test.

Use the comparison test to
determine whether the series which is the sum from π equals one to β of π
cubed plus one over π to the five plus three is convergent or divergent.

We can start by recalling the
comparison test. The comparison test tells us that
for the series which is the sum from π equals one to β of π π and the sum
from π equals one to β of π π, where π π and π π are both greater than
or equal to zero for all π. That if firstly the sum from π
equals one to β of π π converges, and π π is less than or equal to π π
for all π, then the sum from π equals one to β of π π converges. And secondly, if the sum from π
equals one to β of π π diverges, and π π is greater than or equal to π
π for all π, then the sum from π equals one to β of π π also
diverges.

Now, before we use the comparison
test on our series, it may be easier to split it up. We can say that the sum from π
equals one to β of π cubed plus one over π to the five plus three is equal
to the sum from π equals one to β of π cubed over π to the five plus three
plus one over π to the five plus three. And then, we can split this further
by splitting it into two separate series, which will look like this.

We now have two series which we can
perform the comparison test to. Letβs start with the second of this
series. So, thatβs the sum from π equals
one to β of one over π to the power of five plus three. And we can, in fact, compare this
to a π series. A π series is a series which is of
the form of the sum from π equals one to β of one over π to the π. And the π series will diverge if
π is less than or equal to one and converge if π is greater than one.

We can see that our series looks
quite similar to the π series where π is equal to five. So, thatβs the sum from π equals
one to β of one over π to the five. We can apply the comparison test to
these two series. Since in both of these series,
weβre summing from π equals one to β, this tells us that π is always
greater than or equal to one. Looking at the π π and the π π
from our series, when π is greater than or equal to one, π π and π π are both
greater than or equal to zero. Therefore, weβve satisfied the
first condition of the comparison test.

Next, we need to check whether our
sum from π equals one to β of one over π to the five converges or
diverges. Since this is a π series where the
value of π is equal to five, and five is greater than one, this tells us that this
series converges. Therefore, in order to use the
comparison test, we need to satisfy the condition in the first part, which is that
π π is less than or equal to π π for all π. So, thatβs that one over π to the
power of five plus three is less than or equal to one over π to the power of
five.

Since π to the power of five plus
three is greater than π to the power of five for all π greater than or equal to
one, this tells us that this condition is, in fact, satisfied. Therefore, weβve satisfied the last
condition of the comparison test. And we can say that the sum from π
equals one to β of one over π to the five plus three converges. Now, we can focus on the other
series. Thatβs the sum from π equals one
to β of π cubed over π to the five plus three.

Letβs consider π π. So, thatβs π cubed over π to the
power of five plus three. And we can say that this must be
less than or equal to π cubed over π to the power of five since the denominator of
the fraction on the left is bigger than the dominator of the fraction on the
right. Now, π cubed over π to the power
of five is simply one over π squared. Therefore, this implies that we can
compare the series on the left with the series which is the sum from π equals one
to β of one over π squared. And this series is again a π
series where π is equal to two. Two is greater than one. And therefore, this series must
converge.

Here, we have again satisfied all
the conditions for the first part of the comparison test. We can, therefore, come to the
conclusion that the sum from π equals one to β of π cubed over π to the
power of five plus three also converges. We have now found two convergent
series. Therefore, the sum of the two
convergent series will also be convergent. Hence, we have that the sum from π
equal one to β of π cubed plus one ever π to the five plus three is
convergent.

Next, letβs note another common
type of series used in the comparison tests. These are geometric series. These are series which are of the
form of the sum from π equals one to β of π multiplied by π to the power
of π minus one. Now, these geometric series
converge if the absolute value of π is less than one and diverge if the absolute
value of π is greater than or equal to one. Letβs now look at an example of how
we can use geometric series in the comparison test.

Use the comparison test to decide
whether the series which is the sum from π equals one to β of 237 over π
plus 1.1 to the π is convergent or divergent.

Letβs start by recalling the
comparison test. The comparison test tells us that
for the series which is the sum from π equals one to β of π π, and the sum
from π equals one to β of π π, where ππ and π π are both greater than
or equal to zero for all π. That firstly if the sum from π
equals one to β of π π converges, and π π is less than or equal to π π
for all π, then the sum π equals one to β of π π also converges. And secondly, if the sum from π
equals one to β of π π diverges, and π π is greater or equal to π π for
all π, then the sum from π equals one to β of π π also diverges.

Now, our series has been given to
us in the question, which gives that π π is equal to 237 over π plus 1.1 to the
π. Now, it looks as though we may be
able to compare our series to a π series. And this π series is 237
multiplied by the sum from π equals one to β of one over π. Now, for this π series, π is
equal to one. Therefore, this series
diverges. Therefore, weβre looking at the
second part of the comparison test. And we need to try and prove that
π π is greater than or equal to π π for all π. So, thatβs trying to prove that 237
over π plus 1.1 to the π is greater than or equal to 237 over π.

Now, the only differences in these
two fractions is the denominators, in particular the term of 1.1 to the π. Since π is always greater than or
equal to one, therefore, 1.1 to the π is always greater than zero. This tells us that the denominator
of the fraction on the left will be greater than the denominator of the fraction at
the right. And because of this, our inequality
here does not hold, and so we cannot use this π series to apply the comparison
test. Weβll need to try a different
series.

We can, in fact, try a geometric
series. Geometric series is of the form of
the sum of π π to the π. And a geometric series is
convergent if the absolute value of π is less than one and divergent if the
absolute value of π is greater than or equal to one. We now need to find a geometric
series to compare our series to. Looking at our series, we have 1.1
to the π in the denominator. Therefore, letβs try comparing our
series to the sum from π equals one to β of 237 over 1.1 to the π.

For both of these series, since π
is greater than or equal to one, we have that both π π and π π are greater than
or equal to zero for all π. Now, we can rewrite our geometric
series in a form which is easier to find its convergence. And that is the sum from π equals
one to β of 237 multiplied by one over 1.1 to the power of π. Here, we can see that π is equal
to one over 1.1. Since one over 1.1 is less than
one, we have that this geometric series is convergent. Hence, we are looking at the first
part of the comparison test.

We next need to try and prove the
π π is less than or equal to π π for all π. So, that is that 237 over π plus
1.1 to the π is less than or equal to 237 over 1.1 to the π. Since π is greater than or equal
to one, we have that the denominator of π plus 1.1 to the π is greater than the
denominator of 1.1 to the π. And therefore, our inequality
holds. Hence, we have satisfied the
condition of the second part of the comparison test. And so, we reach our conclusion,
which is that the sum from π equals one to β of 237 over π plus 1.1 to the
π is convergent.

We have now covered a variety of
examples of how we can use the comparison test to find convergent or divergent
series. Letβs recap some key points of the
video.

Key Points

The comparison test for series
tells us that for the series which is the sum for π equals one to β of π π
and the sum for π equals one to β of π π, where π π and π π are both
greater than or equal to zero for all π. That firstly if the sum from π
equals one to β of π π is convergent, and π π is less than or equal to π
π for all π, then the sum from π equals one to β of π π is
convergent. And secondly, if the sum from π
equals one to β of π π is divergent, and π π is greater than or equal to
π π for all π, then the sum for π equals one to β of π π is
divergent. When using the comparison test, the
most common types of series we compare to are π series and geometric series.