Video: Alternating Series Test

In this video, we will learn how to determine whether an alternating series is convergent or divergent using the alternating series test.

17:22

Video Transcript

Alternating Series Test

In this video, we’ll learn how to determine whether an alternating series is convergent using the alternating series test. We’ll be looking at a variety of examples of how we can use this convergence test. Let’s start by considering what an alternating series is.

An alternating series is a series which is of the form of the sum from 𝑛 equal to one to ∞ of negative one to the 𝑛 multiplied by π‘Ž 𝑛. This is called an alternating series, since the terms will switch between negative and positive values. This happens because we have a negative one to the 𝑛. And so, when 𝑛 is odd, negative one to the 𝑛 is negative one. And when 𝑛 is even, negative one to the 𝑛 is simply one. Now, an alternating series may come in a slightly different form. We could have the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 plus one π‘Ž 𝑛. And this would still be in alternating series due to the negative one to the power of 𝑛 plus one. However, in this case, when 𝑛 is odd, negative one to the 𝑛 plus one is simply one. And when 𝑛 is even, negative one to the 𝑛 plus one is negative one.

There are other types of series which could be alternating series, which we should look out for. And these are Series, such as these two here. If we look at the series on the left, we have a cos of 𝑛 πœ‹ term. Let’s consider the first few values of cos of 𝑛 πœ‹. For 𝑛 equals one, we have cos of πœ‹, which is equal to negative one. For 𝑛 equals two, we have cos of two πœ‹, which is equal to one. For cos of three πœ‹, we have negative one. And for cos of four πœ‹, we have one. If we carried this on and on and on, we would see the sign of one changing each time we increased 𝑛 by one.

Now, when 𝑛 is an odd number, we have negative one. And when 𝑛 is an even number, we have positive one. Therefore, we can say that cos of 𝑛 πœ‹ is equal to negative one to the 𝑛. And so, we can see that this first series is in fact alternating. Similarly, we can consider sin of two 𝑛 plus one over two πœ‹. We have that sin of three πœ‹ by two is equal to negative one, sin of five πœ‹ by two is equal to one, and sin of seven πœ‹ by two is equal to negative one. And therefore, here, we can also see that sin of two 𝑛 plus one over two πœ‹ is equal to negative one to the 𝑛. Therefore, both of these series are alternating series. And so, we must be careful when identifying an alternating series, since some series may not immediately appear to be an alternating series.

Now, there’s one more condition along with this, which must be satisfied for an alternating series. And that is that we require π‘Ž 𝑛 to be greater than zero for all 𝑛. This is because if the sign of π‘Ž 𝑛 is changing, then our series will no longer alternate between positive and negative between terms. Also, if all of the π‘Ž 𝑛 terms are less than zero, so they’re negative, then we can simply take out a factor of negative one and our series will satisfy these conditions to be an alternating series.

Now that we’ve considered what an alternating series is, let’s have a look at how the alternating series test works. So the alternating series test tells us that for an alternating series, such as the one here, where π‘Ž 𝑛 is greater than zero for all 𝑛, then if firstly, the limits as 𝑛 tends to ∞ of π‘Ž 𝑛 is equal to zero and secondly, the sequence of π‘Ž 𝑛 is a decreasing sequence. Then, the series is a convergent series. Let’s note that this test only tells us if our series converges. If the test does not hold to be true, this does not necessarily mean that our series diverges.

Let’s now look at an example of how this test can be used.

Determine whether the series which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 plus one multiplied by five over 𝑛 plus one minus five over 𝑛 plus two converges or diverges.

Now, looking at our series, we can see that it looks like an alternating series. This is because of the negative one to the 𝑛 plus one term. Now, an alternating series can be of the form as shown here. However, we’re required to that π‘Ž 𝑛 must be greater than zero. In our case, π‘Ž 𝑛 is equal to five over 𝑛 plus one minus five over 𝑛 plus two. We can check whether this is greater than zero by combining the two fractions. We multiply the fraction on the left by 𝑛 plus two over 𝑛 plus two and multiply the fraction on the right by 𝑛 plus one over 𝑛 plus one. After doing this, we’ll have found a common denominator of 𝑛 plus one multiplied by 𝑛 plus two. Then, we can combine the two fractions. This will give five lots of 𝑛 plus two minus five lots of 𝑛 plus one over 𝑛 plus one multiplied by 𝑛 plus two.

Now, we can multiply through the numerator. And for the final step, we can simplify this numerator. What we’re left with is that π‘Ž 𝑛 is equal to five over 𝑛 plus one multiplied by 𝑛 plus two. Now, 𝑛 can be any integer between one and ∞. Therefore, 𝑛 is always positive. And if 𝑛 is always positive, this means that our π‘Ž 𝑛 must also always be positive. Now, we have confirmed that our series is in fact an alternating Series. We can try and use the alternating series test, which tells us that an alternating series converges if the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 is equal to zero and π‘Ž 𝑛 is a decreasing sequence.

Let’s start by finding the limit as 𝑛 tends to ∞ of π‘Ž 𝑛. We have that the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 is equal to the limit as 𝑛 tends to ∞ of five over 𝑛 plus one multiplied by 𝑛 plus two. Now, all of the 𝑛s in our limit are on the denominator of our fraction. And all of these 𝑛s are positive. Therefore, as 𝑛 gets larger and larger and larger, the denominator of this fraction will also get larger. This means that the whole fraction itself will tend towards zero. Hence, we can say that this limit must be equal to zero. Therefore, we’ve satisfied the first condition of the alternating series test.

Next, we need to check whether the sequence π‘Ž 𝑛 is decreasing. If the sequence is decreasing, then π‘Ž 𝑛 must be greater than π‘Ž 𝑛 plus one, since each term will be smaller than the previous one. We can rearrange this inequality. And here, what we have to check is that π‘Ž 𝑛 minus π‘Ž 𝑛 plus one is greater than zero. For π‘Ž 𝑛, we have five over 𝑛 plus one times 𝑛 plus two. Then, for π‘Ž 𝑛 plus one, we have five over 𝑛 plus two times 𝑛 plus three. We can find the difference here by multiplying the fraction on the left by 𝑛 plus three over 𝑛 plus three and the fraction on the right by 𝑛 plus one over 𝑛 plus one.

After combining the two resulting fractions, what we’re left with is five lots of 𝑛 plus three minus five lots of 𝑛 plus one over 𝑛 plus one times 𝑛 plus two times 𝑛 plus three. Simplifying this fraction, what we’re left with is 10 over 𝑛 plus one times 𝑛 plus two times 𝑛 plus three. Since 𝑛 must be between one and ∞, this tells us that 𝑛 itself must be greater than or equal to one. For any 𝑛 value, which is greater than or equal to one, our fraction here, which is π‘Ž 𝑛 minus π‘Ž 𝑛 plus one, must be greater than zero. This is because all of the terms in this fraction will be positive. So we have satisfied the condition required for π‘Ž 𝑛 to be a decreasing sequence. And therefore, the second condition for the alternating series test is also satisfied.

From this, we can come to the conclusion that the series which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 plus one multiplied by five over 𝑛 plus one minus five over 𝑛 plus two converges.

Next, let’s consider what happens if we do not satisfy the conditions of the alternating series test.

Let’s suppose that we have the alternating series, which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 π‘Ž 𝑛. And let’s say that we’ve not satisfied the first condition of the alternating series test. So there are limits as 𝑛 tends to ∞ of π‘Ž 𝑛 is not equal to zero. Now, here, we can not directly say that this series is divergent, since this is not stated within the alternating series test. However, having this piece of information about π‘Ž 𝑛 can be quite useful, especially when we consider the divergence test.

We can recall that the divergent test tells us that for some series, which is the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, if the limit as 𝑛 tends to ∞ of 𝑏 𝑛 is nonzero, then the series diverges. Now, in the case of our alternating series, we have that 𝑏 𝑛 is equal to negative one to the 𝑛 π‘Ž 𝑛. Now, to check whether this alternating series diverges, we need to find the limit of 𝑏 𝑛 as 𝑛 tends to ∞. This is equivalent to the limit as 𝑛 tends to ∞ of negative one to the 𝑛 π‘Ž 𝑛.

Using limit rules, we have that the limit of a product is equal to the product of the limits. And so our limit here is equal to the limit, as 𝑛 tends to ∞ of negative one to the 𝑛 multiplied by the limit as 𝑛 tends to ∞ of π‘Ž 𝑛. Now, the second limit here, which is the limit as 𝑛 tends to ∞ of π‘Ž 𝑛, we have already defined to be nonzero.

Now, there are two cases which could arise here. Maybe, this limit as 𝑛 tends to ∞ of π‘Ž 𝑛 does not exist. If this is the case, then our limit as 𝑛 tends to ∞ of negative one to the 𝑛 π‘Ž 𝑛 does not exist either. This means that the limit as 𝑛 tends to ∞ of 𝑏 𝑛 does not exist. Therefore, it satisfies the condition of the divergence test, since it will not be equal to zero. Therefore, we would be able to say that our series diverges. Now, the other case is that our limit as 𝑛 tends to ∞ of π‘Ž 𝑛 exists and is equal to some constant, which we can call 𝑐, remembering that 𝑐 is not equal to zero. If this is the case, then the limit as 𝑛 tends to ∞ of negative one to the 𝑛 times π‘Ž 𝑛, which is also the limit of 𝑏 𝑛 will be equal to 𝑐 multiplied by the limit as 𝑛 tends to ∞ of negative one to the 𝑛.

Now, as 𝑛 gets larger and larger and larger, negative one to the 𝑛 will keep flipping between negative one and one. Therefore, it will not be approaching any single value. And so, we can say that this limit does not exist. Therefore, the limit as 𝑛 tends to ∞ of 𝑏 𝑛 does not exist either. And so, again, we’ve satisfied the condition of the divergence test. And we can say that our alternating series diverges. Now this can be a pretty useful method for testing divergence of alternating series, if our alternating series fails the first part of the alternating series test.

Let’s now look at another example.

Determine whether the series which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 multiplied by 𝑛 squared plus three over five 𝑛 squared minus one converges or diverges.

Now, our series looks like an alternating series. An alternating series is a series of the form shown here, where π‘Ž 𝑛 is greater than zero for all 𝑛. In our case, π‘Ž 𝑛 is equal to 𝑛 squared plus three over five 𝑛 squared minus one. We can quite clearly see that both the numerator and denominator of the fraction here will be positive for all 𝑛 greater than or equal to one. Therefore, it is true that π‘Ž 𝑛 is greater than zero for all 𝑛. And so, our series is an alternating series. We can, therefore, apply the alternating series test.

This test tells us that a series which of the form of the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 π‘Ž 𝑛, where π‘Ž 𝑛 is greater than zero for all 𝑛, converges. If firstly, the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 is equal to zero. And secondly, if π‘Ž 𝑛 is a decreasing sequence. We can start by finding the limit as 𝑛 tends to ∞ of 𝑛 squared plus three over five 𝑛 squared minus one. We can divide the numerator and denominator by the highest power of 𝑛. So that’s 𝑛 squared. We’re left with the limit as 𝑛 tends to ∞ of one plus three over 𝑛 squared over five minus one over 𝑛 squared.

Now, as 𝑛 tends to ∞, any term with 𝑛 in the denominator. So that’s three over 𝑛 squared and negative one over 𝑛 squared will tend to zero. And we’ll be left with this limit being equal to one-fifth. Now, one-fifth is not equal to zero. Therefore, we have failed to satisfy the first condition of the alternating series test. And so, we can say that this series does not converge by the alternating series test. However, we can now try and apply the divergence test to this series.

The divergence test tells us that if for the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, the limit as 𝑛 tends to ∞ of 𝑏 𝑛 is nonzero, then the series diverges. Now, in our case, 𝑏 𝑛 is equal to negative one to the 𝑛 multiplied by 𝑛 squared plus three over five 𝑛 squared minus one. And this is also equal to negative one to the 𝑛 π‘Ž 𝑛. Therefore, the limit as 𝑛 tends to ∞ of 𝑏 𝑛 is equal to the limit as 𝑛 tends to ∞ of negative one to the 𝑛 π‘Ž 𝑛. And we can split this up using the fact of the limit of the product is equal to the product of the limits. We are left with the limit as 𝑛 tends to ∞ of negative one to the 𝑛 times the limit as 𝑛 tends to ∞ of π‘Ž 𝑛.

Now, we’ve already found the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 and it’s equal to one-fifth. Now, we can consider the limit as 𝑛 tends to ∞ of negative one to the 𝑛. Now, as 𝑛 gets larger and larger and larger, negative one to the 𝑛 will simply be switching between negative one and one. Therefore, we cannot say that this limit tends to a particular value. And therefore, this limit does not exist. Since this limit does not exist, this means that the limit as 𝑛 tends to ∞ of 𝑏 𝑛 also does not exist. Since it does not exist, this limit is therefore not equal to zero. And this does in fact satisfy the condition for the divergence test. Hence, we can come to the conclusion that the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 times 𝑛 squared plus three over five 𝑛 squared minus one diverges.

Next, let’s consider what happens when our sequence of terms π‘Ž 𝑛 is not decreasing for all terms. If we have our alternating series, which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 π‘Ž 𝑛 and we notice that the sequence π‘Ž 𝑛 is not decreasing for all 𝑛. Now, in most cases, there is not a lot we can do here. This series would probably fail the alternating series test and we would not know anything else about it. However, in some cases there may be something we can do.

Let’s suppose that π‘Ž 𝑛 is not decreasing for all 𝑛, but instead is decreasing for 𝑛 is greater than or equal to some capital 𝑁, where capital 𝑁 is simply just some number greater than one. If this is the case, we can rewrite our alternating series. We can rewrite it as the sum from 𝑛 equals one to capital 𝑁 minus one of negative one to the 𝑛 π‘Ž 𝑛 plus the sum from 𝑛 is equal to capital 𝑁 to ∞ of negative one to the 𝑛 π‘Ž 𝑛. Now, we can say that this series on the left is finite as long as each of the terms β€” so that’s negative one to the 𝑛 π‘Ž 𝑛 β€” is itself finite too.

Now, if we consider the series on the right, we have an alternating series for which our sequence of π‘Ž 𝑛 is decreasing. So as long as the limit as 𝑛 goes to ∞ of π‘Ž 𝑛 is equal to zero, then this alternating series satisfies the alternating series test. Therefore, it converges. Now, since the sum of these two series is equal to our original alternating series. If the series on the left is finite and the series on the right converges, then this implies that the series from 𝑛 equals one to ∞ of negative one to the 𝑛 π‘Ž 𝑛 also itself converges. Now, this can be a useful skill to keep in mind when using the alternating series test, especially when our alternating series fails the second condition of the test.

Now that we’ve seen a couple of examples of the alternating series test and a variety of skills which we can use when applying the alternating series test, let’s cover some key point of the video.

Key Points

An alternating series is one of the form of the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 π‘Ž 𝑛 or the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 plus one π‘Ž 𝑛, where π‘Ž 𝑛 is greater than zero for all 𝑛. However, sometimes, negative one to the 𝑛 may appear differently, for example, as cos of 𝑛 πœ‹.

The alternating series test: for an alternating series of the form, which is the sum from 𝑛 equals one to ∞ of negative one to the 𝑛 π‘Ž 𝑛. If firstly, the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 is equal to zero and secondly, if the sequence π‘Ž 𝑛 is a decreasing sequence, then the series converges. The alternating series test only tells us if a series is convergent, not if it is divergent. However, the divergence test can be a useful follow-on if the first condition of the alternating series test fails. So this is if the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 is not equal to zero.

Sometimes, we may be required to split our series if the sequence of π‘Ž 𝑛 is only decreasing for 𝑛 is greater than or equal to capital 𝑁. If this is the case, we can rewrite our series as the sum from 𝑛 equals one to capital 𝑁 minus one of negative one to the 𝑛 π‘Ž 𝑛 plus the sum from 𝑛 is equal to capital 𝑁 to ∞ of negative one to the 𝑛 π‘Ž 𝑛. Now, the time which we’d do this would be if the alternating series test failed on the second condition and our sequence of π‘Ž 𝑛 is only decreasing for 𝑛 is greater than or equal to capital 𝑁. In this case, we’d split up our series like this and potentially have a finite series plus a convergent infinite series, which will tell us that our overall alternating series is convergent.

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