Video: Conditional and Absolute Convergence

In this video, we will learn how to determine if a series is absolutely convergent, conditionally convergent, or divergent.

14:54

Video Transcript

In this video, we will learn how to determine if a series is absolutely convergent, conditionally convergent, or divergent.

We know that if a series is called convergent, it means that the partial sums approach a specific limit. But what does it mean for a series to be called absolutely convergent? We say a series is called absolutely convergent if the series of absolute values is convergent. One thing to notice is that if π‘Ž 𝑛 is a series with positive terms, then the absolute value of π‘Ž 𝑛 is equal to π‘Ž 𝑛. So absolute convergence implies convergence.

First, let’s look at an example where we have to determine whether a series is absolutely convergent.

Is the series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 add one over 𝑛 squared absolutely convergent?

Remember to test for absolute convergence, we need to check whether the series of absolute values is convergent, in other words, is the sum from 𝑛 equals one to ∞ of the absolute value of negative one to the power of 𝑛 add one over 𝑛 squared convergent. First of all, notice that negative one to the power of 𝑛 add one is always going to be either one or negative one, depending on whether the power is even or odd. So if we take the absolute value of negative one raised to the power of 𝑛 add one, this is always going to be one. We know the 𝑛 runs from one to ∞. So 𝑛 squared is always going to be positive. So we can actually rewrite this as one over 𝑛 squared.

Remember that we’re trying to determine whether this converges or diverges. But we actually recognize the sum from 𝑛 equals one to ∞ of one over 𝑛 squared to be a series that we know. It’s a 𝑃-series. So we use the fact that a 𝑃-series converges if 𝑃 is greater than one and diverges if 𝑃 is less than or equal to one. So for our question, we see that this is a 𝑃-series with 𝑃 equal to two. Because this is greater than one, we can say that the sum from 𝑛 equals one to ∞ of one over 𝑛 squared is convergent. So because we found the series of absolute values to be convergent, then the series is absolutely convergent.

Interestingly, if we find a series, which is not absolutely convergent, it may still be convergent. We call this conditional convergence. A series is conditionally convergent if the series is convergent but not absolutely convergent. In other words, the sum from 𝑛 equals one to ∞ of the absolute value of π‘Ž 𝑛 diverges. But the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 converges. And if a series is not absolutely convergent and it’s not conditionally convergent, then it’s divergent.

Let’s see an example of conditional convergence.

Is the alternating harmonic series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 add one multiplied by one over 𝑛 absolutely convergent, conditionally convergent, or divergent?

Let’s firstly remember that a series is absolutely convergent if the series of absolute values is convergent. And a series is conditionally convergent if the series of absolute values diverges but the series converges. And otherwise, the series is divergent. So let’s start by testing for absolute convergence. We can see that negative one raised to the power of 𝑛 add one is always going to give us one when we take the absolute value. So this is in fact the same as the sum from 𝑛 equals one to ∞ of one over 𝑛. But this is actually a series that we’re familiar with. It’s the harmonic series. And we know that the harmonic series diverges. So the alternating harmonic series is not absolutely convergent. But is it conditionally convergent or divergent?

So our next step is to test the alternating harmonic series for convergence. Because that’s an alternating series, we can do this with the alternating series test. Recall this says that for an alternating series, the sum of negative one to the power of 𝑛 add one multiplied by 𝑏 𝑛 if 𝑏 𝑛 is decreasing and the limit as 𝑛 approaches ∞ of 𝑏 𝑛 is equal to zero, then π‘Ž 𝑛 is convergent. So for the alternating harmonic series, we can say that 𝑏 𝑛 equals one over 𝑛. So is 𝑏 𝑛 decreasing? Well, as 𝑛 increases, one over 𝑛 does decrease. So that condition is satisfied. But does the limit as 𝑛 approaches ∞ of 𝑏 𝑛 equal zero? Well, the limit as 𝑛 approaches ∞ of one over 𝑛 is going to be one over ∞, which we know is zero. So that condition is satisfied. So because we found that the alternating harmonic series is not absolutely convergent, but it is convergent, we can conclude that the alternating harmonic series is conditionally convergent.

We can summarize the check for absolute convergence, conditional convergence, and divergence in a helpful diagram. Let’s say we want to find out whether the series π‘Ž 𝑛 is absolutely convergent, conditionally convergent, or divergent. We begin by testing whether the series of absolute values is convergent or divergent. Let’s say that we find that the series of absolute values is convergent. Then, the series π‘Ž 𝑛 is absolutely convergent. But if we find that the series of absolute values is divergent, then the series π‘Ž 𝑛 is not absolutely convergent. But it may still be conditionally convergent. So we try a different test on the series π‘Ž 𝑛 to check for convergence, for example, the alternating series test.

And if we find that the series π‘Ž 𝑛 converges, then we say that the series π‘Ž 𝑛 is conditionally convergent. But if we find that the series π‘Ž 𝑛 diverges, then we conclude that the series π‘Ž 𝑛 is divergent. So these are the three possible conclusions that we can draw.

So let’s now see some more examples.

Consider the series the sum from 𝑛 equals one to ∞ of sin of 𝑛 over 𝑛 cubed. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

Recall that a series π‘Ž 𝑛 is absolutely convergent if the series of absolute values is convergent. And if we find that the series is not absolutely convergent, it may still be conditionally convergent. So we then test the series for convergence or divergence. So let’s begin by testing this series for absolute convergence. So we want to find out whether the sum from 𝑛 equals one to ∞ of the absolute value of sin of 𝑛 over 𝑛 cubed is convergent or divergent.

Well, because 𝑛 only runs through positive values from one to ∞, 𝑛 cubed is always going to be positive. So this is just the sum from 𝑛 equals one to ∞ of the absolute value of sin of 𝑛 over 𝑛 cubed. Now we know that sin of 𝑛 will always be between negative one and one. So we can say that the absolute value of sin of 𝑛 will always be less than or equal to one, which means that we can write the absolute value of sin of 𝑛 over 𝑛 cubed is less than or equal to one over 𝑛 cubed. Writing it this way allows us to do a direct comparison. Recall that this means if π‘Ž 𝑛 is less than 𝑏 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 also converges.

And one over 𝑛 cubed is actually a series we recognize. Recall that a 𝑃-series is a series of the form the sum for 𝑛 equals one to ∞ of one over 𝑛 to the 𝑃 power. And this converges if 𝑃 is greater than one and diverges if 𝑃 is less than or equal to one. So one over 𝑛 cubed is a 𝑃-series with 𝑃 equal to three. So one over 𝑛 cubed converges. So by direct comparison, the absolute value of sin of 𝑛 over 𝑛 cubed also converges. Then, because we found the series of absolute values to be convergent, then our series the sum from 𝑛 equals one to ∞ of sin of 𝑛 over 𝑛 cubed is absolutely convergent.

State whether the series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 add one multiplied by two over the square root of 𝑛 add one converges absolutely, conditionally, or not at all.

Firstly, recall that for a series π‘Ž 𝑛, this is absolutely convergent if the series of absolute values converges. And it’s conditionally convergent if the series of absolute values diverges. But the series itself converges. So let’s first of all find out whether this series is absolutely convergent or not. This means testing whether the series from 𝑛 equals one to ∞ of the absolute value of negative one to the power of 𝑛 add one multiplied by two over the square root of 𝑛 add one is convergent or divergent.

Well, negative one to the power of 𝑛 add one is always going to be one or negative one. But if we take the absolute value, it will always be one. Whereas two over the square root of 𝑛 add one will always be positive because 𝑛 runs through positive values. So we can write this as the sum from 𝑛 equals one to ∞ of two over the square root of 𝑛 add one. Then, we can use the constant multiplication rule to bring the two to the front of the sum. From here, we need to work out whether this series converges or diverges. One way we can actually do this is with a direct comparison with the harmonic series.

Because for 𝑛 is greater than two, we have that one over 𝑛 is less than one over the square root of 𝑛 add one. And we know that if we have π‘Ž 𝑛 less than 𝑏 𝑛 where π‘Ž 𝑛 diverges, then 𝑏 𝑛 also diverges. And we know that the sum from 𝑛 equals one to ∞ of one over 𝑛 is the harmonic series which diverges. Then, the sum from 𝑛 equals one to ∞ of one over the square root of 𝑛 add one also diverges. So we found that the series of absolute values diverges, which means that this series is not absolutely convergent. But it could still be conditionally convergent. So we’re going to test the series itself for convergence. So let’s clear some space.

We can firstly bring the constant two to the front of the sum. And then, if we look at this negative one to the power of 𝑛 add one, this creates an alternating series because it makes the terms alternate between positive and negative. So we can decide whether this series is convergent or divergent using the alternating series test. Remember that this says for a series π‘Ž 𝑛, where π‘Ž 𝑛 is equal to negative one to the power of 𝑛 add one multiplied by 𝑏 𝑛, if 𝑏 𝑛 is decreasing and the limit as 𝑛 approaches ∞ of 𝑏 𝑛 is equal to zero, then the series π‘Ž 𝑛 is convergent. So for our series 𝑏 𝑛 is equal to one over the square root of 𝑛 add one. But is this decreasing? Well, for 𝑏 𝑛 to be decreasing, we need 𝑏 𝑛 to be greater than 𝑏 𝑛 add one.

Well, we can see that as 𝑛 increases by one, the square root of 𝑛 add one is going to get bigger. So one over the square root of 𝑛 add one is going to decrease. So 𝑏 𝑛 is decreasing. Then, we need to check whether the limit as 𝑛 approaches ∞ of 𝑏 𝑛 is equal to zero. In other words, is the limit as 𝑛 approaches ∞ of one over the square root of 𝑛 add one zero? Well, the square root of 𝑛 add one is increasing as 𝑛 gets bigger. So this will be one over ∞. So as 𝑛 approaches ∞, then one over the square root of 𝑛 add one approaches one over ∞. And so the limit as 𝑛 approaches ∞ is zero. So both of these conditions are satisfied. So this series is convergent.

Remember that we said that a series is conditionally convergent if the series of absolute values diverges but the series converges. And that’s exactly what we’ve found here. The series of absolute values was divergent. But we found the series itself to be convergent. So we can conclude that this series converges conditionally.

But why is that helpful to differentiate between absolutely convergent and conditionally convergent series? Well, absolutely convergent infinite series holds some of the same properties as finite sums. For example, if we have a finite sum, then any rearrangement of the terms still gives the same sum. And this also holds true for an absolutely convergent series. Any rearrangement yields the same sum. But this is not true for a conditionally convergent series. This is because rearranging the terms of a series changes the partial sums. So this can change the limit of the partial sums when some of the terms are negative. So we don’t have that issue with absolutely convergent series. But of course, this doesn’t apply to conditionally convergent series.

As an example, the alternating harmonic series, which we’ve seen is convergent, can be shown to converge to the natural log of two. But if the terms in the series are rearranged so that every positive term is followed by two negative terms, this does change the value of the sum. So with a conditionally convergent series, rearrangement changes the relative rate at which positive and negative terms are used and in turn changes the sum of the series. In fact, we can actually use this to rearrange a conditionally convergent series to converge to any value we want. But it’s beyond the scope of this video to go through these proofs in detail.

Let’s now summarize some of the main points. A series π‘Ž 𝑛 is absolutely convergent if the series of absolute values is convergent. And a series π‘Ž 𝑛 is conditionally convergent if the series of absolute values is divergent, but the series itself is convergent. And finally, if π‘Ž 𝑛 is an absolutely convergent series with sum 𝑠, then any rearrangement of π‘Ž 𝑛 yields the same sum 𝑠.

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