### Video Transcript

In this video, weβre going to learn
how to define a power series and how these differ from previous series we might have
come across. Weβll also learn how to find the
radius of convergence and the interval of convergence of these sides of series using
the ratio or root test.

Up until this stage, youβll likely
have worked with infinite geometric series, harmonic and π series, and alternating
series. Now we look to define a new type of
series, which we call a power series. Power series are useful in analysis
since they arise as Taylor series of infinitely differentiable functions. They can be of the form shown. Itβs the sum of π π times π₯ to
the πth power for values of π between zero and β. Here π₯ is a variable, whilst the
π πs are the coefficient of our series.

Consider now what happens if we
define a fixed value for π₯. For example, let π₯ be equal to
one. In this series, we obtain a series
of constants, which we can then test for convergence or divergence using our usual
tests.

Now itβs important to realise that
a power series may converge for some values of π₯ and diverge for others. We also define the sum of the
series as the function π of π₯, which is equal to π naught plus π one π₯, and so
on. Whose domain is the set of all π₯
for which the series converges. We might notice that π looks a lot
like a polynomial, though we should also observe that it is slightly different from
one since it has an infinite number of terms.

We might observe what happens when
we take π π to be equal to one. The power series becomes the
geometric series one plus π₯ plus π₯ squared plus π₯ cubed all the way up to π₯ to
the πth power. Which we recall converges when the
absolute value of π₯ is less than one and diverges when the absolute value of π₯ is
greater than or equal to one. Letβs look to generalise.

A series of the form the sum of π
π times π₯ minus π to the πth power for values of π between zero and β is called
a power series in π₯ minus π or a power series centred at π or about π. Notice that weβve written π naught
for the term when π equals zero. So we are assuming that π₯ minus π
to the power of zero is equal to one even when π₯ equals π. Notice also that if π₯ equals π,
all terms are zero for π is greater than or equal to one. So the power series always
converges when π₯ is equal to π. And we can use ideally the ratio or
root tests to test for convergence of these series. Letβs see what that might look
like.

For what values of π₯ is the series
the sum of π factorial times π₯ plus five to the πth power for values of π
between zero and β convergent?

Letβs recall that we can use the
ratio test to test for convergence. This says that suppose we have the
series to find as the sum of π π. Then if the limit as π approaches
β of the absolute value of π π plus one over π π is less than one, the series is
absolutely convergent. If the converse is true, if the
limit is greater than one, then the series is divergent. And if the limit is equal to the
one, the series may be divergent, conditionally convergent, or absolutely
convergent.

Now since weβre looking to find the
values of π₯ for which the series is convergent, weβre interested in this first case
where the limit is less than one. And so we define π π to be equal
to π factorial times π₯ plus five to the πth power. And this means that π π plus one
is π plus one factorial times π₯ plus five to the power of π plus one.

Weβre looking to find when the
limit as π approaches β of the absolute value of the quotient of these is less than
one. So thatβs π plus one factorial
times π₯ plus five to the power of π plus one over π factorial times π₯ plus five
to the power of π. We begin by recalling that when
weβre dividing two numbers whose base is the same, we can simply subtract the
exponents. And this means that π₯ plus five to
the power of π plus one divided by π₯ plus five to the power of π is simply π₯
plus five to the power of one or just π₯ plus five.

Then we recall that π plus one
factorial is equal to π plus one times π times π minus one times π minus two,
and so on. Well, thatβs the same as π plus
one times π factorial. So we can divide through by π
factorial. And so we have the limit as π
approaches β of the absolute value of π plus one times π₯ plus five. π₯ plus five is independent of
π. So we can take the absolute value
of π₯ plus five outside of our limit.

And then we notice that the limit
as π approaches β of the absolute value of π plus one must itself be β. The absolute value of π₯ plus five
multiplied by β will always be greater than one, except if the absolute value of π₯
plus five itself is equal to zero. Now since the absolute value of π₯
plus five we know is equal to zero, we donβt actually need the absolute value
signs. And so we solved by subtracting
five from both sides. And we see that π₯ is equal to
negative five. The series given will therefore
only converge when π₯ is equal to negative five. Letβs now look to generalise the
convergence of a power series.

For a given power series the sum of
π π times π₯ minus π to the πth power, there are only three possibilities. The first is when the series
converges only when π₯ is equal to π. The second is when the series
converges for all values of π₯. Or thirdly, there exists a positive
number capital π
such that the series converges if the absolute value of π₯ minus
π is less than π
and diverges if the absolute value of π₯ minus π is greater than
π
. We call this capital π
the radius
of convergence of the power series.

By convention, for part one, we say
that π
is equal to zero. And for part two, we say that π
is
equal to β. Then weβre able to define the
interval of convergence of a power series as the interval that consists of all
values of π₯ for which the series converges. In our first possibility, the
interval just consists of the single point at π, whereas in our second possibility,
itβs from negative β to β. In our third, we rewrite the
absolute value of π₯ minus π being less than π
as π₯ minus π is greater than
negative π
and less than π
.

Now when π₯ itself is an end point
of the interval, in other words, when π₯ is equal to π plus or minus capital π
,
the series might converge or diverge at one or both end points. And so the ratio and root test will
always fail when π₯ is an end point of the interval of convergence.

In these cases, we need to check
the end points using some alternative test. Letβs have a look at how to find
the radius of convergence and interval of convergence of a power series.

Consider the power series the sum
of three π₯ to the πth power over π plus five for π equals zero to β. Find the interval of convergence of
the power series and find the radius of convergence of the power series.

To test for convergence of our
power series, weβre going to recall the ratio test. The part of the ratio test weβre
interested in says that, given a series the sum of π π, if the limit as π
approaches β of the absolute value of π π plus one over π π is less than one,
then the series is absolutely convergent. If we look at our series, we see
that we can define π π as three π₯ to the πth power over π plus five. And then π π plus one is three π₯
to the power of π plus one over π plus one plus five.

And then of course that denominator
simplifies nicely to π plus six. Our job is to establish where the
limit as π approaches β of the absolute value of the quotient of these is less than
one. And we know that when dividing by a
fraction, we multiply by the reciprocal of that very same fraction. So the inside of our limit becomes
three π₯ to the power of π plus one over π plus six times π plus five over three
π₯ to the πth power. And we now see that we can divide
through by three π₯ to the πth power.

Okay, so now we have the limit as
π approaches β of the absolute value of three π₯ times π plus five over π plus
six. We notice that three π₯ is
independent of π. So we can take the absolute value
of three π₯ outside our limit. And then we divide through both the
numerator and denominator of the fraction by π. Thatβs the highest power of π in
our denominator. And so the denominator becomes π
divided by π, which is one, plus five over π. And then the numerator becomes one
plus six over π. And as π approaches β, five over
π and six over π approach zero. And so the limit as π approaches β
of the absolute value of one plus five over π over one plus six over π is simply
the absolute value of one or one.

So we need to establish where the
absolute value of three π₯ times one is less than one, or simply where the absolute
value of three π₯ is less than one. Since three is a purely positive
number, we can divide through both sides of our inequality by three without
affecting those absolute value signs. And we find that the absolute value
of π₯ must be less than one-third. So the series converges if the
absolute value of π₯ is less than one-third. And conversely, we can say it will
diverge if the absolute value of π₯ is greater than one-third. And in fact, weβve found the radius
of convergence of our power series. Itβs π
equals one-third.

Now another way of writing the
absolute value of π₯ being less than one-third is saying that π₯ must be greater
than negative one-third and less than a third. And so weβve determined an interval
of convergence. We certainly know that the power
series converges for these values of π₯. But we donβt know what happens at
the end points of our interval. Weβre going to clear some space and
substitute π₯ equals negative one-third and π₯ equals one-third into our original
power series. And weβll see if these series
converge or diverge using an alternative test.

Weβll begin by letting π₯ be equal
to negative one-third. Then our series is the sum from π
equals zero to β of three times negative one-third to the πth power over π plus
five. Which simplifies to the sum from π
equals zero to β of negative one to the πth power over π plus five. And you might recognise this. Itβs an alternating series. This negative one to the πth power
indicates so.

So weβre going to use the
alternating series test to establish whether this converges or diverges. This series is to be used when we
have a series the sum of π π, where π π is negative one to the πth power times
π π or negative one to the power of π plus one times π π. And here π π must be greater than
or equal to zero for all π. Then if the limit as π approaches
β of π π equals zero and π π is a decreasing sequence, then we can say our
series the sum of π π is convergent.

Letβs rewrite our series as the sum
from π equals zero to β of negative one to the πth power times π plus five to the
power of negative one. We now see itβs of the form given,
where π is equal to π plus five to the power of negative one. Letβs check the limit as π
approaches β of this π π is indeed zero.

Well, we know itβs equal to the
limit as π approaches β of one over π plus five. And as π π gets larger, π plus
five, the denominator of the fraction, gets larger and larger. So one over π plus five gets
smaller and smaller and eventually approaches zero. So the limit of π π as π
approaches β is indeed zero.

Next, we need to check whether π
π is a decreasing sequence. To do so, letβs differentiate the
entire function with respect to π. Using the general power rule, we
see thatβs negative π plus five to the power of negative two, which is negative one
over π plus five all squared. For all values of π between zero
and β, the denominator of our fraction is always positive. So negative one over a positive
number gives us a negative number, which therefore means that we do indeed have a
decreasing sequence.

Weβve satisfied all the criteria
for our series. And we can therefore say that the
series when π₯ is equal to negative one-third converges. Weβre now going to clear some space
and test π₯ equals a third. This time, our series is the sum
from π equals zero to β of three times a third to the πth power over π plus five,
which is one to the πth power over π plus five, which is of course simply one over
π plus five.

Now this is actually an example of
a general harmonic series. And we of course know that these
diverge. Our series therefore diverges for
values of π₯ greater than or equal to negative one-third and less than
one-third. And our interval is shown.

In our final example, weβll
consider how we can use the root test to find convergence.

Consider the power series the sum
from π equals zero to β of π times two π₯ to the πth power. Determine the radius of convergence
of the power series and determine the interval of convergence of the power
series.

To test for convergence of this
power series, we could use the ratio or the root test. Letβs look at the root test. The part of the test weβre
interested in says suppose we have a series the sum of π π. If the limit as π approaches β of
the πth root of the absolute value of π π, which can of course be alternatively
written as the absolute value of π π to the power of one over π, is less than
one. Then the series is absolutely
convergent and hence convergent.

So we define π π for our series
to be π times two π₯ to the πth power. Weβre looking to establish where
the limit as π approaches β of the absolute value of π times two π₯ to the πth
power to the power of one over π is less than one. We know that finding an πth root
doesnβt change the sign. So we can rewrite this and say that
the limit as π approaches β of the absolute value of π to the power of one over π
times two π₯ to the πth power to the power of one over π must be less than
one. Two π₯ to the πth power to the
power of one over π is simply two π₯. And so our limit is as shown.

We know that two π₯ is independent
of π. So we can take the absolute value
of two π₯ outside of our limit. The problem we have now is that if
we try to evaluate the limit as π approaches β of the absolute value of π to the
power of one over π. We get the absolute value of β to
the power of one over β, which is β to the power of zero. And we know thatβs indeterminate
form. But we might recall that we can
write π to the power of one over π as π to the power of the natural log of π
over π. And we know that as π approaches
β, the natural logarithm of π over π approaches zero. And so the limit as π approaches β
of the absolute value of π to the power of one over π is π to the power of zero,
which is simply one.

Of course we donβt need the
absolute value signs for one. And so we see weβre interested in
when the absolute value of two π₯ is less than one. Since two is a purely positive
number, we can divide through. And we see that the absolute value
of π₯ is less than one-half. These are the values of π₯ for
which the series converges. And we can therefore say that the
radius convergence of our power series is one-half.

Another way of representing the
absolute value of π₯ being less than one-half is to say that π₯ must be greater than
negative one-half and less than one-half. And so weβve determined an interval
for convergence. But we do need to check if the
power series converges or diverges at the end points of our interval. In other words, when π₯ is equal to
negative one-half or π₯ is equal to one-half.

Weβre going to plug these into our
original power series and see if those series converge or diverge using alternative
tests. Letβs clear some space. Weβll begin by letting π₯ equal
negative one-half. The series is the sum from π
equals zero to β of π times two minus negative one-half to the πth power. And this becomes π times negative
one to the power of π.

To establish whether this series
converges or diverges, we could use the alternating series test with π π equals
π. Since the limit as π approaches β
of π π is not equal to zero, we can say that this series ultimately diverges. So letβs try when π₯ is equal to
one-half. We have the sum from π equals zero
to β of π times two times a half to the πth power. Thatβs π times one to the πth
power. And of course one to the πth power
will always be one. So we have the sum from π equals
zero to β of π.

This time, we can actually apply
the πth term divergence test. The limit as π approaches β of π
is β. Thatβs not equal to zero. And so, once again, when π₯ is
equal to a half, our series diverges. And so the interval of convergence
of our power series is the open interval negative one-half to one-half.

In this video, we saw that a series
of the form the sum of π π times π₯ minus π to the πth power for values of π
between zero and β is called a power series about π. We also saw that we can use the
ratio test and root test to find the radius of convergence and interval of
convergence of our power series. But that itβs also important that
we should test the end points of the interval for convergence or divergence.