### Video Transcript

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.