### Video Transcript

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.