Video: The 𝑛th Term Divergence Test

In this video, we will learn how to apply the nth term test for divergence which states that a series diverges if the limit of its nth term does not approach zero.

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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 π‘Žπ‘› from 𝑛 equals one to infinity is convergent, then the limit as 𝑛 approaches infinity of π‘Žπ‘› 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 π‘Žπ‘› 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 π‘Žπ‘› 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 π‘Žπ‘› 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 π‘Žπ‘› does not exist or if the limit is not equal to zero, then the series the sum of π‘Žπ‘› 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 π‘Žπ‘› 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 π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯. And that’s provided the limits exist and the limit of 𝑔 of π‘₯ 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 π‘Žπ‘› 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 π‘Žπ‘› does not exist or the limit is not equal to zero, the series the sum of π‘Žπ‘› 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 π‘Žπ‘› 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 π‘Žπ‘› 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 π‘Žπ‘› is not equal to zero or does not exist, then the series the sum of π‘Žπ‘› 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 π‘Žπ‘› 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’HΓ΄pital’s rule. This says if the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to infinity over infinity, then the limit as π‘₯ approaches π‘Ž prime of π‘₯ over 𝑔 prime of π‘₯ will tell us the value of the limit as π‘₯ approaches π‘Ž of 𝑓of π‘₯ over 𝑔 of π‘₯. 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 π‘Žπ‘› is not equal to zero or does not exist, then the series the sum of π‘Žπ‘› from 𝑛 equals one to infinity is divergent. We saw that if the series the sum of π‘Žπ‘› from 𝑛 equals one to infinity is convergent, then the limit as 𝑛 approaches infinity of π‘Žπ‘› 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 π‘Ž of 𝑛 is equal to zero, we can’t conclude that the series is convergent. And in fact, we say that the test fails.

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