### Video Transcript

In this video, weβre going to learn
about something called the πth term divergence test. Weβll learn how to use it to prove
that a series diverges and look at what happens when the test fails. Weβll also consider what this test
can tell us about a convergent series.

We begin by recalling that a
sequence is called convergent if the limit of the sum of its first π terms as π
approaches infinity is equal to some finite number π , known as the sum of the
series. Similarly, the sequence is
divergent if the converse is true, if the sum of the series does not have a finite
limit. So how do we test for this?

Well, there are a number of
methods. In this video, weβre going to use
something called the πth term divergence test to establish whether a series
diverges or whether the test fails. The first definition weβre
interested in is that if the series the sum of ππ from π equals one to infinity
is convergent, then the limit as π approaches infinity of ππ is equal to
zero. And itβs really important to
realise that the converse of this theorem is not necessarily true.

If the limit as π approaches
infinity of ππ is equal to zero, we canβt conclude that the series is
convergent. For example, the harmonic series
the sum of one over π, we have ππ equals one over π, which approaches zero as π
approaches infinity. But that is a divergent series.

And so we come to the πth term
divergence test. This says if the limit as π
approaches infinity does not exist or is not equal to zero, then the series the sum
of ππ from π equals one to infinity is divergent.

Notice that, once again, this shows
us that if the limit is equal to zero, we know nothing about the convergence or
divergence of this series. Itβs really important that we
realise that when the πth term divergence test gives an answer of zero, we say that
the test fails. Letβs now consider an example.

Using the πth term test, determine
whether the series the sum of π over π squared plus one from π equals zero to
infinity is divergent or the test fails.

Remember, the πth term test for
divergence says that if the limit as π approaches infinity of ππ does not exist
or if the limit is not equal to zero, then the series the sum of ππ from π equals
one to infinity is divergent. We also recall that if the limit is
equal to zero, we canβt tell whether the series converges or diverges and we say
that the test fails.

Now notice the sum in our question
is from π equals zero to infinity, as opposed to from π equals one to
infinity. In practice, this doesnβt really
matter. Weβre looking for absolute
divergence, essentially whatβs happening to the series as π gets larger and
larger. And if we look carefully, we see
that when π equals zero, our first term is also zero. So we can split this into zero plus
the sum from π equals one.

In our case then, weβre going to
let ππ be equal to π over π squared plus one. And weβre going to evaluate the
limit as π approaches infinity of π over π squared plus one. And when evaluating a limit, we
should always check whether we can use direct substitution. In this case, if we were to
substitute π equals infinity into the expression, we get infinity over infinity,
which we know to be indeterminate. So instead, we look for a way to
manipulate the expression π over π squared plus one. We do so by dividing both the
numerator and denominator by π squared.

Remember, we can do this as it
creates an equivalent fraction. And we choose π squared as itβs
the highest power of π in our denominator. So we get the limit as π
approaches infinity of π over π squared over π squared over π squared plus one
over π squared. This simplifies to the limit as π
approaches infinity of one over π over one plus one over π squared.

Weβre then going to use the
division law for limits. This says that the limit as π₯
approaches π of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of
π of π₯ over the limit as π₯ approaches π of π of π₯. And thatβs provided the limits
exist and the limit of π of π₯ is not equal to zero. And so this becomes the limit as π
approaches infinity of one over π over the limit as π approaches infinity of one
plus one over π squared.

And we can now substitute π equals
infinity into this expression. As π gets larger, one over π gets
smaller. Ultimately, it approaches zero. Similarly, as π gets larger, one
over π squared also approaches zero. And so our limit becomes zero over
one plus zero, which is of course simply zero. And so since the limit as π
approaches infinity of ππ is equal to zero, we see that the test fails.

Weβll now consider another
example.

What can we conclude by applying
the πth term divergence test in the series the sum of three π over the square root
of six π squared plus four π plus five for values of π from one to infinity?

Remember, the πth term test for
divergence says that if the limit as π approaches infinity of ππ does not exist
or the limit is not equal to zero, the series the sum of ππ from π equals one to
infinity is divergent. And we recall that if the limit is
equal to zero, we canβt tell whether the series converges or diverges and we say
that the test fails.

In this question, weβre going to
let ππ be equal to three π over the square root of six π squared plus four π
plus five. And so we need to evaluate the
limit of this expression as π approaches infinity. We canβt use direct
substitution. If we were to do so, we would get
infinity over infinity, which we know is indeterminate. So instead, we need to find a way
to manipulate our expression and see if that will help us to evaluate the limit.

To make our life just a little bit
easier, letβs apply the constant multiple rule. And this is that the limit as π
approaches infinity of some constant times the function in π is equal to that
constant times the limit of the function in π. So essentially, we can take the
constant three outside of our limit.

And then this next step is going to
appear a little bit strange. Weβre going to divide both the
numerator and denominator of our expression by π. The numerator then becomes one. And the denominator becomes one
over π times the square root of six π squared plus four π plus five. Weβre now going to take this one
over π inside our square root. To do that, weβll need to square
it. So the denominator becomes the
square root of six π squared over π squared plus four π over π squared plus five
over π squared. And then, actually, this simplifies
quite nicely.

We now have three times the limit
as π approaches infinity of one over the square root of six plus four over π plus
five over π squared. And weβre now ready to apply direct
substitution. As π gets larger, four over π and
five over π squared get smaller. They approach zero. One and six are both independent of
π. So our limit becomes one over the
square root of six.

We want to rationalise the
denominator here. So we multiply both the numerator
and denominator of our fraction by the square root of six. And we see that the limit as π
approaches infinity of ππ is three times the square root of six over six, which is
root six over two. We notice that this is not equal to
zero. And this leads us to conclude that
the series the sum of three π over the square root of six π squared plus four π
plus five between π equals one and infinity is divergent.

Weβll now consider a slightly more
complicated example that requires use of an additional rule for finding limits.

What can we conclude by applying
the πth term divergence test in the series the sum of two times the natural log of
π over three π for π equals one to infinity?

We begin by recalling that the πth
term test for divergence says that if the limit as π approaches infinity of ππ is
not equal to zero or does not exist, then the series the sum of ππ from π equals
one to infinity is divergent. And indeed, if that limit is equal
to zero, we canβt tell whether the series converges or diverges and we say that the
test fails.

In our question then, weβre going
to let ππ be equal to two times the natural log of π over three π. And so our job is to evaluate the
limit as π approaches infinity of this expression. If we were to simply apply direct
substitution, then weβd find that our limit is equal to infinity over infinity. And of course, thatβs
indeterminate.

So instead, weβre going to recall
LβHΓ΄pitalβs rule. This says if the limit as π₯
approaches π of π of π₯ over π of π₯ is equal to infinity over infinity, then the
limit as π₯ approaches π of π prime of π₯ over π prime of π₯ will tell us the value of
the limit as π₯ approaches π of π of π₯ over π of π₯. We can also use this formula if our
limit is equal to zero over zero. But weβre not interested in that
case.

Now of course, weβre working with
π. So weβre going to need to
differentiate two times the natural log of π and three π with respect to π. The derivative of the natural log
of π is one over π. So when we differentiate two times
the natural log of π with respect π, we get two over π. And then the derivative of three π
is simply three. So we can now evaluate this as π
approaches infinity.

As π gets larger, two over π gets
smaller. And as π approaches infinity
therefore, two over π approaches zero. We find that this is therefore
equal to zero over three, which is zero. And we find that the test fails or
itβs inconclusive.

In this video, weβve seen that the
πth term divergence test can tell us whether a series is divergent. And it tells us that if the limit
as π approaches infinity of ππ is not equal to zero or does not exist, then the
series the sum of ππ from π equals one to infinity is divergent. We saw that if the series the sum
of ππ from π equals one to infinity is convergent, then the limit as π
approaches infinity of ππ will be zero. But that itβs really important to
realise that the converse of this theorem is not necessarily true. If the limit as π approaches
infinity of π of π is equal to zero, we canβt conclude that the series is
convergent. And in fact, we say that the test
fails.