### Video Transcript

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.