### Video Transcript

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.