Video Transcript
In this video, we’re going to learn
about something called the 𝑛th term divergence test. We’ll learn how to use it to prove
that a series diverges and look at what happens when the test fails. We’ll 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 𝑛 terms as 𝑛
approaches infinity is equal to some finite number 𝑠, 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, we’re going to use
something called the 𝑛th term divergence test to establish whether a series
diverges or whether the test fails. The first definition we’re
interested in is that if the series the sum of 𝑎𝑛 from 𝑛 equals one to infinity
is convergent, then the limit as 𝑛 approaches infinity of 𝑎𝑛 is equal to
zero. And it’s really important to
realise that the converse of this theorem is not necessarily true.
If the limit as 𝑛 approaches
infinity of 𝑎𝑛 is equal to zero, we can’t conclude that the series is
convergent. For example, the harmonic series
the sum of one over 𝑛, we have 𝑎𝑛 equals one over 𝑛, which approaches zero as 𝑛
approaches infinity. But that is a divergent series.
And so we come to the 𝑛th term
divergence test. This says if the limit as 𝑛
approaches infinity does not exist or is not equal to zero, then the series the sum
of 𝑎𝑛 from 𝑛 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. It’s really important that we
realise that when the 𝑛th term divergence test gives an answer of zero, we say that
the test fails. Let’s now consider an example.
Using the 𝑛th term test, determine
whether the series the sum of 𝑛 over 𝑛 squared plus one from 𝑛 equals zero to
infinity is divergent or the test fails.
Remember, the 𝑛th term test for
divergence says that if the limit as 𝑛 approaches infinity of 𝑎𝑛 does not exist
or if the limit is not equal to zero, then the series the sum of 𝑎𝑛 from 𝑛 equals
one to infinity is divergent. We also recall that if the limit is
equal to zero, we can’t tell whether the series converges or diverges and we say
that the test fails.
Now notice the sum in our question
is from 𝑛 equals zero to infinity, as opposed to from 𝑛 equals one to
infinity. In practice, this doesn’t really
matter. We’re looking for absolute
divergence, essentially what’s happening to the series as 𝑛 gets larger and
larger. And if we look carefully, we see
that when 𝑛 equals zero, our first term is also zero. So we can split this into zero plus
the sum from 𝑛 equals one.
In our case then, we’re going to
let 𝑎𝑛 be equal to 𝑛 over 𝑛 squared plus one. And we’re going to evaluate the
limit as 𝑛 approaches infinity of 𝑛 over 𝑛 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 𝑛 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 𝑛 over 𝑛 squared plus one. We do so by dividing both the
numerator and denominator by 𝑛 squared.
Remember, we can do this as it
creates an equivalent fraction. And we choose 𝑛 squared as it’s
the highest power of 𝑛 in our denominator. So we get the limit as 𝑛
approaches infinity of 𝑛 over 𝑛 squared over 𝑛 squared over 𝑛 squared plus one
over 𝑛 squared. This simplifies to the limit as 𝑛
approaches infinity of one over 𝑛 over one plus one over 𝑛 squared.
We’re then going to use the
division law for limits. This says that the limit as 𝑥
approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of
𝑓 of 𝑥 over the limit as 𝑥 approaches 𝑎 of 𝑔 of 𝑥. And that’s provided the limits
exist and the limit of 𝑔 of 𝑥 is not equal to zero. And so this becomes the limit as 𝑛
approaches infinity of one over 𝑛 over the limit as 𝑛 approaches infinity of one
plus one over 𝑛 squared.
And we can now substitute 𝑛 equals
infinity into this expression. As 𝑛 gets larger, one over 𝑛 gets
smaller. Ultimately, it approaches zero. Similarly, as 𝑛 gets larger, one
over 𝑛 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 𝑛
approaches infinity of 𝑎𝑛 is equal to zero, we see that the test fails.
We’ll now consider another
example.
What can we conclude by applying
the 𝑛th term divergence test in the series the sum of three 𝑛 over the square root
of six 𝑛 squared plus four 𝑛 plus five for values of 𝑛 from one to infinity?
Remember, the 𝑛th term test for
divergence says that if the limit as 𝑛 approaches infinity of 𝑎𝑛 does not exist
or the limit is not equal to zero, the series the sum of 𝑎𝑛 from 𝑛 equals one to
infinity is divergent. And we recall that if the limit is
equal to zero, we can’t tell whether the series converges or diverges and we say
that the test fails.
In this question, we’re going to
let 𝑎𝑛 be equal to three 𝑛 over the square root of six 𝑛 squared plus four 𝑛
plus five. And so we need to evaluate the
limit of this expression as 𝑛 approaches infinity. We can’t 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, let’s apply the constant multiple rule. And this is that the limit as 𝑛
approaches infinity of some constant times the function in 𝑛 is equal to that
constant times the limit of the function in 𝑛. 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. We’re going to divide both the
numerator and denominator of our expression by 𝑛. The numerator then becomes one. And the denominator becomes one
over 𝑛 times the square root of six 𝑛 squared plus four 𝑛 plus five. We’re now going to take this one
over 𝑛 inside our square root. To do that, we’ll need to square
it. So the denominator becomes the
square root of six 𝑛 squared over 𝑛 squared plus four 𝑛 over 𝑛 squared plus five
over 𝑛 squared. And then, actually, this simplifies
quite nicely.
We now have three times the limit
as 𝑛 approaches infinity of one over the square root of six plus four over 𝑛 plus
five over 𝑛 squared. And we’re now ready to apply direct
substitution. As 𝑛 gets larger, four over 𝑛 and
five over 𝑛 squared get smaller. They approach zero. One and six are both independent of
𝑛. 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 𝑛
approaches infinity of 𝑎𝑛 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 𝑛 over the square root of six 𝑛 squared plus four 𝑛
plus five between 𝑛 equals one and infinity is divergent.
We’ll now consider a slightly more
complicated example that requires use of an additional rule for finding limits.
What can we conclude by applying
the 𝑛th term divergence test in the series the sum of two times the natural log of
𝑛 over three 𝑛 for 𝑛 equals one to infinity?
We begin by recalling that the 𝑛th
term test for divergence says that if the limit as 𝑛 approaches infinity of 𝑎𝑛 is
not equal to zero or does not exist, then the series the sum of 𝑎𝑛 from 𝑛 equals
one to infinity is divergent. And indeed, if that limit is equal
to zero, we can’t tell whether the series converges or diverges and we say that the
test fails.
In our question then, we’re going
to let 𝑎𝑛 be equal to two times the natural log of 𝑛 over three 𝑛. And so our job is to evaluate the
limit as 𝑛 approaches infinity of this expression. If we were to simply apply direct
substitution, then we’d find that our limit is equal to infinity over infinity. And of course, that’s
indeterminate.
So instead, we’re going to recall
L’Hôpital’s rule. This says if the limit as 𝑥
approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to infinity over infinity, then the
limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥 will tell us the value of
the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥. We can also use this formula if our
limit is equal to zero over zero. But we’re not interested in that
case.
Now of course, we’re working with
𝑛. So we’re going to need to
differentiate two times the natural log of 𝑛 and three 𝑛 with respect to 𝑛. The derivative of the natural log
of 𝑛 is one over 𝑛. So when we differentiate two times
the natural log of 𝑛 with respect 𝑛, we get two over 𝑛. And then the derivative of three 𝑛
is simply three. So we can now evaluate this as 𝑛
approaches infinity.
As 𝑛 gets larger, two over 𝑛 gets
smaller. And as 𝑛 approaches infinity
therefore, two over 𝑛 approaches zero. We find that this is therefore
equal to zero over three, which is zero. And we find that the test fails or
it’s inconclusive.
In this video, we’ve seen that the
𝑛th term divergence test can tell us whether a series is divergent. And it tells us that if the limit
as 𝑛 approaches infinity of 𝑎𝑛 is not equal to zero or does not exist, then the
series the sum of 𝑎𝑛 from 𝑛 equals one to infinity is divergent. We saw that if the series the sum
of 𝑎𝑛 from 𝑛 equals one to infinity is convergent, then the limit as 𝑛
approaches infinity of 𝑎𝑛 will be zero. But that it’s really important to
realise that the converse of this theorem is not necessarily true. If the limit as 𝑛 approaches
infinity of 𝑎 of 𝑛 is equal to zero, we can’t conclude that the series is
convergent. And in fact, we say that the test
fails.
