### Video Transcript

Find the interval of convergence
for the power series the sum from π equals one to β of π₯ plus three to the πth
power divided by six π.

The question wants us to find the
interval of convergence for this power series. And we recall that an interval of
convergence for a power series is an interval which contains all of the values of π₯
such that the series converges. To help us find the interval of
convergence of this power series, weβre going to use the ratio test. And we recall the ratio test tells
us if we have an infinite series the sum of π π, where the limit as π approaches
β of the absolute value of the ratio of successive terms is less than one. Then our series converges
absolutely. However, if our series the sum of
π π has the limit as π approaches β of the absolute value of the ratio of
successive terms is greater than one, then the ratio test tells us that our series
must be divergent.

In our case, our series is the sum
from π equals one to β of π₯ plus three to the πth power divided by six π. So weβll set our sequence π π
equal to this and then attempt to use the ratio test. Using this gives us the limit as π
approaches β of the absolute value of the ratio of successive terms is equal to the
limit as π approaches β of the absolute value of π π plus one divided by π
π. Where π π plus one is equal to π₯
plus three to the power of π plus one divided by six times π plus one. And π π is equal to π₯ plus three
to the πth power divided by six π.

We can simplify this limit. Instead of dividing by the fraction
π₯ plus three to the πth power divided by six π, we can multiply by the reciprocal
of this fraction. This gives us the limit as π
approaches β of the absolute value of π₯ plus three to the power of π plus one
divided by six times π plus one multiplied by six π divided by π₯ plus three to
the πth power. We can simplify this limit by
canceling the shared factor of six in our numerator and our denominator and
canceling π of the shared factors of π₯ plus three in our numerator and our
denominator. This leaves us with the limit as π
approaches β of the absolute value of π₯ plus three times π divided by π plus
one.

Since our limit is as π is
approaching β, we can see that π₯ plus three remains constant. Itβs invariant with respect to
π. So we can take this constant
outside of our limit, where we must be careful to take the absolute value of π₯ plus
three since we were calculating the limit of an absolute value. We now need to evaluate the limit
as π approaches β of the absolute value of π divided by π plus one. Thereβs a few different ways of
doing this. For example, we could divide π by
π plus one by using algebraic division. However, weβre going to use the
method of dividing both our numerator and our denominator by the highest power of π
which appears in the fraction.

We can see the only power of π
which appears in our fraction is π itself, so weβll divide both our numerator and
our denominator by π. Dividing the numerator of π by π
just gives us one. And dividing our denominator of π
plus one through by π will give us one plus one divided by π. So this gives us the absolute value
of π₯ plus three multiplied by the limit as π approaches β of the absolute value of
one divided by one plus one over π.

Weβre now ready to evaluate this
limit. Since our limit is as π is
approaching β, the one in our numerator and the one in our denominator remain
constant. However, as π is approaching β,
one divided by π is approaching zero. This means our limit is approaching
the absolute value of one divided by one, which is just equal to one. Therefore, weβve shown the limit as
π approaches β of the absolute value of the ratio of successive terms of the series
given to us in the question is equal to the absolute value of π₯ plus three.

And remember the ratio test tells
us if this value is less than one, then our series is convergent. And if this value is greater than
one, then the series is divergent. So the ratio test tells us that our
series will converge if the absolute value of π₯ plus three is less than one. And saying the absolute value of π₯
plus three is less than one is the same as saying that π₯ plus three is greater than
negative one and π₯ plus three is less than one. We can then subtract three from all
parts of this inequality to see that π₯ must be bigger than negative four and π₯
must be less than negative two.

Similarly, using the ratio test, we
can conclude if the absolute value of π₯ plus three is greater than one, then our
series will be divergent. And using a similar argument, we
can show that this is the same as saying π₯ is less than negative four or π₯ is
greater than negative two. But remember, the interval of
convergence contains all of the values of π₯ such that our series is convergent. And weβve looked at all of the
values of π₯ except when π₯ is equal to negative four and when π₯ is equal to
negative two since at these two values of π₯, the ratio test was inconclusive. We found the limit as π approached
β of the absolute value of the ratio of successive terms for these two values of π₯
was just equal to one.

To help us determine whether the
series converges or diverges in these two cases, weβre going to clear some space and
then substitute the values of π₯ is equal to negative four and π₯ is equal to
negative two directly into our power series. Weβll start when π₯ is equal to
negative two. This gives us the sum from π
equals one to β of negative two plus three to the πth power divided by six π.

We can start by taking the constant
factor of one-sixth outside of our sum. Next, we can evaluate negative two
plus three to just give us one. And we know that one to the πth
power is always equal to one. And weβve now shown that our series
is equal to one-sixth multiplied by the harmonic series, which we know diverges. Therefore, our series was divergent
when π₯ was equal to negative two. And so this wonβt be in our
interval of convergence.

We can do the same when π₯ is equal
to negative four. We can simplify our series to give
us one-sixth multiplied by the sum from π equals one to β of negative one to the
πth power divided by π. We can now see that our summand
negative one to the πth power divided by π is an alternating sequence. And one over π is positive and
decreasing. Therefore, by the alternating
series test, we know that this series is convergent. Therefore, weβve shown all of the
values of π₯ where our series is convergent is when π₯ is greater than negative four
and less than negative two and when π₯ is equal to negative four.

Therefore, weβve shown that the
interval of convergence for the power series the sum from π equals one to β of π₯
plus three to the πth power divided by six π is when π₯ is greater than or equal
to negative four and when π₯ is less than negative two.