Video: Infinite Geometric Series

In this video, we will learn how to determine whether a geometric series is convergent and find its limit.

16:45

Video Transcript

Infinite Geometric Series

In this video, we will learn how to determine whether a geometric series is convergent and, if so, to find its value. The geometric series is an important example of an infinite series. You may come across this type of series when dealing with physical processes, such as the height of a bouncing ball, or in other areas of maths, such as fractal geometry. Geometric series can be written in the following form using sigma notation. We would read this mathematical statement as the sum from 𝑛 equals one to infinity of π‘Ž times π‘Ÿ to the power of 𝑛 minus one. Now, you might also see this represented as the sum from 𝑛 equals zero to infinity of π‘Ž times π‘Ÿ to the power of 𝑛.

You should know that this is just an equivalent form but we’re in an index shift. For this video, the definition that we’ll choose to work with is the sum which starts at 𝑛 equals one. Okay, so the general form of the geometric series will be the sum of the following terms. Remember, since this is an infinite series we’ll have an infinite number of terms. You’ll note that we can characterize our geometric series using two things. We have our first term, which is π‘Ž, and we have a common ratio, which is π‘Ÿ. Looking at our series, we see that each successive term can be obtained by multiplying the previous term by the common ratio. This is the characteristic of all geometric series. It’s also worth noting that as a direct result of this characteristic, we could find the common ratio by dividing any term by the previous term. Let us now consider an example geometric series to give us some context.

Imagine we had a geometric series where the first term π‘Ž was equal to three and the common ratio π‘Ÿ was equal to a half. The terms in our example series would, of course, start with the first term, which is a three. And we would multiply it by the common ratio, which is a half to obtain each successive term. Of course, this pattern would continue for infinitely many terms. Using our sigma notation, we could express this series in the following way: the sum from 𝑛 equals one to infinity of three times a half to the power of 𝑛 minus one. Okay, in general, when working with a series, we would like to be able to determine the value of the series. If a series is divergent, of course, we cannot assign it to finite value. However, if a series is convergent, we can. But in practice, it is sometimes very difficult to determine this value. For geometric series, however, we have a useful formula that can be used. Let us work towards finding this.

We’ll first need to get a couple of cases out of the way. Consider the case where our first term π‘Ž is equal to zero and our common ratio π‘Ÿ takes any value. Since our first term is zero and we multiply it by the common ratio to get successive terms, this would mean all of our terms would be zero. Of course, there is no much sense in creating a sum of zeros. And so from here onwards, we’ll assume that the trivial case of π‘Ž equal zero will be ignored for all the values of π‘Ÿ that we explore. What about the case when our common ratio π‘Ÿ is equal to one. Here, π‘Ž is not equal to zero but takes any finals value. Well, obviously our first term is π‘Ž one. Since each successive term is obtained by multiplying the previous term by one, all of our terms would be the same as the first term.

To move forward, Let us now think back to the concept of a partial sum. If the partial sum, 𝑆 𝑛, is defined as the sum of the first 𝑛 terms in a series, then the partial some for the geometric series that we’re considering with a common ratio of one would be 𝑛 times π‘Ž one. Since all of our terms are the same, we’ll simply multiply the first term π‘Ž one by the number of terms that we’re something. We can now use this result in conjunction with a common technique for finding the value of an infinite series. And that is taking the limit as 𝑛 approaches infinity of the partial sum. Now, it should be fairly easy to convince ourselves that as 𝑛 approaches infinity, the limit will also approach a positive or negative infinity, depending on the sign of π‘Ž one. And of course, this is a particular way of expressing that the limit does not exist.

Here, we have somewhat loosely proved that if the common ratio of a geometric series is equal to one, then we cannot find its value. And hence, the series is divergent. Now, we could extend this conclusion a little bit further with some logic. What if our common ratio π‘Ÿ was greater than one? This would mean that each successive term would get larger and larger in magnitude. Again, it should be fairly easy to convince ourselves that if this happens, our series will be divergent. So a geometric series with a common ratio π‘Ÿ greater than one is also divergent. And we cannot assign it to value. Okay, enough about the series that we can’t assign a value to. What about those that we can?

For this, let’s shift focus onto our partial sum 𝑆 𝑛. We can write out the terms to our partial sum in the familiar way. However, since this is not an infinite sum, we will have a final term, which is π‘Ž times π‘Ÿ to the power of 𝑛 minus one. Our next step will be to multiply our partial sum by the common ratio π‘Ÿ. And you’ll see why we’ve done this in a moment. Doing this will mean each of our previous terms will be multiplied by π‘Ÿ. We have a near identical set of terms, but we can imagine they have each been shifted along one place. We no longer have a term at the beginning, which is just π‘Ž on its own. But we have gained a term at the end, which is π‘Ž times π‘Ÿ to the power of 𝑛. If we imagine our first line as equation one and our second line as equation two, let’s see what happens when we subtract two from one, so 𝑆 𝑛 minus π‘Ÿ times 𝑆 𝑛.

Since all of our middle terms are identical, they cancel out. What we’re left with is π‘Ž minus π‘Ž times π‘Ÿ to the power of 𝑛. We can factorize this equation and then we can divide by one minus π‘Ÿ. Great, we have obtained a useful result for the partial sum up to 𝑛 terms. But what about an infinite sum, such as the infinite series that we are considering? Let’s clear some room to continue. Let us use the same technique as earlier taking the limit as 𝑛 goes to infinity of the partial sum. Okay, this may be a bit of a lead, but let’s consider the following. If our common ratio π‘Ÿ is greater than negative one but less than one, then as 𝑛 approaches infinity, π‘Ÿ to the power of 𝑛 will approach zero. This is because we are multiplying a number which is less than one in magnitude by itself multiple times. So it will get smaller and smaller.

We can express this in limit form as the limit as 𝑛 goes to infinity of π‘Ÿ to the power of 𝑛 is equal to zero. Remember, this is only true in the specific case where this inequality is satisfied. It is actually more common to see this inequality written as the magnitude or the absolute value of π‘Ÿ is less than one. Okay, let’s return to our calculations and apply what we’ve discovered. The first thing we can do is to factor out π‘Ž divided by one minus π‘Ÿ from our limit. Next, we know that as 𝑛 approaches infinity π‘Ÿ to the power of 𝑛 approaches zero, assuming that the absolute value of π‘Ÿ is less than one. This means that we can reexpress our limit as one minus zero. Of course, one minus zero is just one. What we’ve just found is that, in this particular case, when the absolute value of π‘Ÿ is less than one, the limit as 𝑛 goes to infinity of the partial sum is equal to π‘Ž divided by one minus π‘Ÿ.

Remember, we were using this limit to find the value of our geometric series. Since the limit exists and is finite, this means that our geometric series is convergent. And its value will also be π‘Ž divided by one minus π‘Ÿ. Note that if instead we had considered the case where the absolute value of π‘Ÿ was greater than or equal to one, we would have found the time limit did not exist and hence our geometric series would be divergent. Okay, we’ve now reached our conclusion.

So let’s summarize this information to make it easier to process. Here, we have a general representation of a geometric series for the first term of π‘Ž and a common ratio of π‘Ÿ. The series is convergent if the absolute value of the common ratio π‘Ÿ is less than one. And the value of the sum in this case is π‘Ž divided by one minus π‘Ÿ. If instead, the absolute value of π‘Ÿ is greater than or equal to one, the series is divergent. Okay, We’ve been through a lot here, but hopefully this final section has distilled down the important information. Let’s now look through some examples to see how we can apply our knowledge.

Is the series 884 plus 884 divided by nine plus 884 divided by 81 and so on convergent or divergent?

For this question, we have been given a series. Now, we have not actually been given very many terms. But if we look at what we have been given, it seems that each successive term can be found by multiplying the previous term by one over nine. This is the characteristic of a geometric series. The general representation for a geometric series is shown here. This is the sum from 𝑛 equals one to infinity of π‘Ž times π‘Ÿ to the power of 𝑛 minus one. a geometric series can be characterized by the first term π‘Ž and the common ratio π‘Ÿ. Again, we know that successive terms can be found by multiplying the previous term by the common ratio. And we could extend this logic to say that the common ratio could be found by dividing any term by the term that comes before it.

If we try and match the series given in our question to the general form of a geometric series, we can see that we have a first term π‘Ž of 884. And we have a common ratio π‘Ÿ of one over nine. If we wanted to express the series given by our question using the sigma notation, we would replace the π‘Ž with 884 and the π‘Ÿ with one over nine from our general representation of a geometric series. Now, when working with geometric series, we can use the following rule. If the absolute value of the common ratio π‘Ÿ is less than one, the series is convergent. And if the absolute value of π‘Ÿ is greater than or equal to one, this series is divergent. Now, we have a common ratio π‘Ÿ of one over nine, and we clearly see that the absolute value of one over nine is less than one. This allows us to conclude that the series given by our question is convergent. With this logic, we have answered the question and we have concluded that the given series is convergent.

Let us now move on to another example.

Find the common ratio of an infinite geometric sequence given that the sum is 52 and the first term is 14.

For this question, we have been asked to find the common ratio of an infinite geometric sequence. The first thing we should note is that sequences and series are closely interlinked. We have been told the sum of our sequence is 52. At this point, we remember that the sum of an infinite sequence is what defines an infinite series. In essence, we’ll need to use the tools for an infinite geometric series for this question. The first thing we can do is to recall the general form for an infinite geometric series. This type of series can be characterized by a first term π‘Ž and a common ratio π‘Ÿ for which successive terms are found by multiplying the previous term by the common ratio.

Now, a general rule that we can use for geometric series is if the absolute value of the common ratio π‘Ÿ is less than one, then the series is convergent. If this is the case, then the value of the sum is equal to the first term π‘Ž divided by one minus π‘Ÿ, the common ratio. If instead the absolute value of π‘Ÿ is greater than or equal to one, then the series is divergent. And of course, for a divergence series we cannot assign this sum a value. Now returning back to our question, the first thing we can note is that the question has indeed assigned the sum a finite value. This implies the series is convergent, and we can ignore the case of a divergent series. It also means that our sum can be expressed as π‘Ž divided by one minus π‘Ÿ. And of course, this is equal to 52.

The other piece of information that we’ve been given by our question is that the first term is 14. This is great since we can use the given value for π‘Ž to find the unknown value for π‘Ÿ using our equation. We first substitute π‘Ž equals 14. We now solve for π‘Ÿ, which is the common ratio that the question is asking for. We can multiply both sides by one minus π‘Ÿ, multiply both the terms in parentheses by 52. We then subtract 52 from both sides of our equation. And for our final step, we can divide both sides by negative 52. Doing so, we find the value of π‘Ÿ, the common ratio, is 19 divided by 26. With this step, we have answered our question. We used the given information and our knowledge of geometric series and how these relate to geometric sequences to find that the common ratio is 19 over 26.

As a quick side note, if the question had asked us to express the series, we could have done so using sigma notation, as shown here. Let’s now move on to an example where we’re asked to find the value of a series.

Does the series the sum from 𝑛 equals one to infinity of three times one over 10 to the power of 𝑛 minus one converge or diverge? If it converges, find the value of the series.

The first thing we should recognize for this question is that we have been given an infinite geometric series. The general form of this type of question is shown here using sigma notation as with this question. Note that this type of series is characterized by a first term π‘Ž and a common ratio π‘Ÿ. A general rule that we use for geometric series is that if the absolute value of the common ratio is less than one, then the series is convergent. And the value of the sum is equal to the first term π‘Ž divided by one minus the common ratio π‘Ÿ. If instead the absolute value of the common ratio is greater than or equal to one, the series is divergent and we cannot assign its sum a value. Okay, this is all interesting, but let’s see how it applies to our question.

If we look at our question, we can see that the series that we’ve been given exactly matches the general form for a geometric series that we should be familiar with. We have a first term π‘Ž of three. And we have a common ratio π‘Ÿ of one over 10. Note that the index of 𝑛 equals one matches on both of our sums, as does the exponents on our common ratio 𝑛 minus one. If the index number had not matched, we might need to perform an index shift. And if the exponents had no matched, we might need to perform some factorization. Lucky for us, this is not the case, and so we continue, safe in the knowledge that our common ratio is one over 10.

To move forward, our first step is quite simple. We observed that the absolute value of our common ratio π‘Ÿ is less than one. This allows us to conclude that the series is convergent. It also tells us that we can find the value of our sum using this formula. If we apply this to our series, its value is π‘Ž divided by one minus π‘Ÿ. Of course, substituting π‘Ž equals three and π‘Ÿ equals one over 10, we obtain three divided by one minus one over 10. This is the same as three divided by nine over 10 which is three times 10 over nine. Or if we cancel the common factor of three on the top and bottom of our fraction, we’re left with 10 over three. If we tidy up our working, we see that we have now answered the question. Using our knowledge of infinite geometric series, we concluded that the series given by the question is convergent and has a value of 10 over three.

Now, before we finish, this question has an interesting characteristic which is worth mentioning. Let us look back at our series and write out the terms. Of course, our first term is three. We can obtain our next term by multiplying by the common ratio one over 10 to get three over 10. Continuing this pattern, we get three over 100 three over 1000 and so on. Now, if we were to represent these fractions in decimal form, we might begin to see a pattern emerging. Since we started with a three and multiplied by one over 10 for each successive term, we will have a three in every position following the decimal point. And when we say every position, this is really true. Since we’re dealing with an infinite series, we will never run out of terms. And so our threes really do repeat forever.

In fact, what we have created here is a recurring decimal, specifically 3.3 recurring. Now, this is very interesting to us, since we have essentially represented our infinite geometric series as a recurring decimal. We won’t go into too much detail for this video, but suffice to say, the reverse process could also be conducted. If we were given a recurring decimal, we could represent it as an infinite geometric series. Okay, but why do this? Well, we’ve just shown that our series is convergent on that we can find its value. The value we found was the convenient fraction of 10 over three. It, therefore, follows that our infinite geometric series is equal to 3.3 recurring, which is also equal to 10 over three. So to conclude, it might not seem like it at first, but geometric series give us a way to represent recurring decimals. If we extend this concept a little bit further, they also give us a way to express recurring decimals as fractions, which is sometimes much more convenient.

Okay, to finish of this video, let us go through some key points. The general form of an infinite geometric series is as follows. We would read this as the sum from 𝑛 equals one to infinity of π‘Ž times π‘Ÿ to the power of 𝑛 minus one. You might also come across the alternative form, which is the sum from 𝑛 equals zero to infinity of π‘Ž times π‘Ÿ to the power of 𝑛. A geometric series can be characterized by π‘Ž first term π‘Ž and a common ratio π‘Ÿ. Successive terms of a geometric series can be found by multiplying the previous term by the common ratio π‘Ÿ.

If the absolute value of the common ratio π‘Ÿ is less than one, then the series is convergent. In this case, the value of the series is finite and is equal to the first term π‘Ž divided by one minus the common ratio π‘Ÿ. If instead, the absolute value of the common ratio is greater than or equal to one, then the series is divergent. And of course in this case, we cannot find its value. Infinite geometric series also give us a way to represent recurring decimals and a way to express them as a fraction. But note, we didn’t explore this particular concept into much detail for this video. So perhaps, it’s a topic which you might want to investigate further.

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