### 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 π equals one to infinity of π times π to the power of
π minus one. Now, you might also see this
represented as the sum from π equals zero to infinity of π times π to the power
of π.

You should know that this is just
an equivalent form but weβre in an index shift. For this video, the definition that
weβll choose to work with is the sum which starts at π 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 weβll have an infinite number of terms. Youβll note that we can
characterize our geometric series using two things. We have our first term, which is
π, and we have a common ratio, which is π. 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. Itβs 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 was equal to three and the common ratio π 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 π equals one to infinity of
three times a half to the power of π 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.

Weβll first need to get a couple of
cases out of the way. Consider the case where our first
term π is equal to zero and our common ratio π 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, weβll
assume that the trivial case of π equal zero will be ignored for all the values of
π that we explore. What about the case when our common
ratio π is equal to one. Here, π is not equal to zero but
takes any finals value. Well, obviously our first term is
π 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, π π, is
defined as the sum of the first π terms in a series, then the partial some for the
geometric series that weβre considering with a common ratio of one would be π times
π one. Since all of our terms are the
same, weβll simply multiply the first term π one by the number of terms that weβre
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 π
approaches infinity of the partial sum. Now, it should be fairly easy to
convince ourselves that as π approaches infinity, the limit will also approach a
positive or negative infinity, depending on the sign of π 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 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 greater than one is also divergent. And we cannot assign it to
value. Okay, enough about the series that
we canβt assign a value to. What about those that we can?

For this, letβs shift focus onto
our partial sum π π. 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 times π to the power of π
minus one. Our next step will be to multiply
our partial sum by the common ratio π. And youβll see why weβve done this
in a moment. Doing this will mean each of our
previous terms will be multiplied by π. 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 on its own. But we have gained a term at the
end, which is π times π to the power of π. If we imagine our first line as
equation one and our second line as equation two, letβs see what happens when we
subtract two from one, so π π minus π times π π.

Since all of our middle terms are
identical, they cancel out. What weβre left with is π minus π
times π to the power of π. We can factorize this equation and
then we can divide by one minus π. Great, we have obtained a useful
result for the partial sum up to π terms. But what about an infinite sum,
such as the infinite series that we are considering? Letβs clear some room to
continue. Let us use the same technique as
earlier taking the limit as π goes to infinity of the partial sum. Okay, this may be a bit of a lead,
but letβs consider the following. If our common ratio π is greater
than negative one but less than one, then as π approaches infinity, π to the power
of π 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 π goes to infinity of π to the power of π 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 is less than
one. Okay, letβs return to our
calculations and apply what weβve discovered. The first thing we can do is to
factor out π divided by one minus π from our limit. Next, we know that as π approaches
infinity π to the power of π approaches zero, assuming that the absolute value of
π 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 weβve just found is that, in
this particular case, when the absolute value of π is less than one, the limit as
π goes to infinity of the partial sum is equal to π divided by one minus π.

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
divided by one minus π. Note that if instead we had
considered the case where the absolute value of π 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, weβve now reached our
conclusion.

So letβs 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 and a common ratio of
π. The series is convergent if the
absolute value of the common ratio π is less than one. And the value of the sum in this
case is π divided by one minus π. If instead, the absolute value of
π is greater than or equal to one, the series is divergent. Okay, Weβve been through a lot
here, but hopefully this final section has distilled down the important
information. Letβs 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 π equals one
to infinity of π times π to the power of π minus one. a geometric series can be
characterized by the first term π and the common ratio π. 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 of 884. And we have a common ratio π of
one over nine. If we wanted to express the series
given by our question using the sigma notation, we would replace the π with 884 and
the π 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 is less than one, the series is convergent. And if the absolute value of π is
greater than or equal to one, this series is divergent. Now, we have a common ratio π 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, weβll 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 and a common ratio π 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 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 divided by one minus π, the common
ratio. If instead the absolute value of π
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 divided by one minus π. And of course, this is equal to
52.

The other piece of information that
weβve been given by our question is that the first term is 14. This is great since we can use the
given value for π to find the unknown value for π using our equation. We first substitute π equals
14. We now solve for π, which is the
common ratio that the question is asking for. We can multiply both sides by one
minus π, 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,
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. Letβs now move on to an example
where weβre asked to find the value of a series.

Does the series the sum from π
equals one to infinity of three times one over 10 to the power of π 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 and a common ratio π. 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 divided by one minus the common ratio π. 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
letβs see how it applies to our question.

If we look at our question, we can
see that the series that weβve been given exactly matches the general form for a
geometric series that we should be familiar with. We have a first term π of
three. And we have a common ratio π of
one over 10. Note that the index of π equals
one matches on both of our sums, as does the exponents on our common ratio π 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 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 divided by one minus π. Of course, substituting π equals
three and π 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, weβre 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 weβre 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 wonβt 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, weβve 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
π equals one to infinity of π times π to the power of π minus one. You might also come across the
alternative form, which is the sum from π equals zero to infinity of π times π to
the power of π. A geometric series can be
characterized by π first term π and a common ratio π. Successive terms of a geometric
series can be found by multiplying the previous term by the common ratio π.

If the absolute value of the common
ratio π 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 divided by one minus the common
ratio π. 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 didnβt explore this
particular concept into much detail for this video. So perhaps, itβs a topic which you
might want to investigate further.