### Video Transcript

Find the interval of convergence
for the power series the sum from π equal zero to β of π₯ to the power of three π
plus two all divided by five π plus 10.

The question asks us to find the
interval of convergence for this power series. We recall that we call πΌ an
interval of the real numbers, the interval of convergence for a series of ππ. If π₯ is in the interval πΌ, then
the sum of ππ converges. And if π₯ is not in the interval
πΌ, then the sum of ππ must diverge. So weβre looking for which values
of π₯ will cause the sum from π equals zero to β of π₯ to the power of three π
plus two all divided by five π plus 10 to converge.

One of the best ways to find an
interval of convergence for a power series is to try using the ratio test. We recall that the ratio test tells
us that if we let πΏ be equal to the limit as π approaches β of the absolute value
of ππ plus one divided by π, that is the limit as π approaches β of the absolute
value of the ratio of successive terms. If this value of πΏ is less than
one, then we can conclude the sum to β of ππ converges. And if πΏ is greater than one, then
the sum to β of ππ diverges.

So letβs apply the ratio test to
the power series given to us in the question. We have the limit as π approaches
β of the absolute value of ππ plus one divided by ππ which is equal to the limit
as π approaches β of our absolute value of our ππ plus one term. Which is equal to π₯ to the power
of three multiplied by π plus one plus two all divided by five multiplied by π
plus one plus 10. And from this, we need to divide
our term ππ, which is π₯ to the power of three π plus two all divided by five π
plus 10.

Now, letβs start expanding the
parentheses in our first term. We get three multiplied by π plus
one plus two can be simplified to three π plus five. And the denominator five multiplied
by π plus one plus 10 can be simplified to just five π plus 15. And we recall that instead of
dividing by our term ππ, we can multiply it by the reciprocal of ππ. We now recall one of our rules for
exponents, which says that π₯ to the power of π divided by π₯ to the power π is
just equal to π₯ to the power of π minus π.

So we can simplify the shared
factors of π₯ in the numerator and the denominator by just changing our numerator to
have π₯ to the power of three π plus five minus three π minus two. Since three π plus five minus
three π minus two is just equal to three, we have the limit as π approaches β of
the absolute value of π₯ cubed multiplied by five π plus 10 all divided by five π
plus 15. At this point, since weβre taking a
limit as π approaches β, we can notice that the value of π₯ is not dependent on the
value of π. So π₯ is a constant with respect to
π, and we can take the value of π₯ cubed out of our limit.

Well, we must be careful that it
must be the absolute value of π₯ cubed since weβre taking the limit of an absolute
value. This means the new limit that we
have to evaluate is the limit as π approaches β of the absolute value of five π
plus 10 all divided by five π plus 15. One way of evaluating limits like
this is to divide both the numerator and the denominator by π. Evaluating this gives us the
absolute value of π₯ cubed multiplied by the limit as π approaches β of the
absolute value of five plus 10 over π all divided by five plus 15 over π.

Since our limit is as π is
approaching β, both 10 divided by π and 15 divided by π are getting smaller and
smaller, and so their limit is equal to zero. Therefore, when we evaluate this
limit, we get the absolute value of five divided by five, which is just equal to
one. And so our limit as π approaches β
of the absolute value of ππ plus one divided by ππ is just equal to the absolute
value of π₯ cubed. We now recall that our ratio test
says that if this limit is less than one, then we can conclude our power series
converges. And if this limit is greater than
one, then we can conclude our power series diverges.

The first thing we notice is that
the absolute value of π₯ cubed being less than one is exactly the same as saying
that negative one is less than π₯ cubed is less than one. One way of finding the region of π₯
where this is true is to sketch a graph of π¦ equals π₯ cubed. We find that π₯ cubed is equal to
one when π₯ is equal to one and π₯ cubed is equal to negative one when π₯ is equal
to negative one. And we see that our region when our
curve π¦ equals π₯ cubed is between one and negative one is when π₯ is between one
and negative one. So what this means is that when π₯
is between negative one and one, we have that our limit πΏ is less than one, and so
our power series must converge.

Similarly, from the sketch, we can
see that if π₯ is bigger than one or if π₯ is less than negative one, then our
output values of π₯ cubed are no longer between one and negative one. And so the absolute value of π₯
cubed must be bigger than one. Therefore, if π₯ is bigger than
one, or if π₯ is less than negative one, then the size of our limit πΏ is bigger
than one. And so by the ratio test, we have
that our power series must be divergent. So far, what we have shown is that
our interval of convergence for our power series must at least include the values
from negative one to one. So this leaves us with the
question, βAre one and negative one in our interval of convergence, which we will
call πΌ?β

Letβs go back to the series given
to us in the question. Letβs start when π₯ is negative
one. We see that our power series is
equal to the sum from π equals zero to β of negative one to the power of three π
plus two all divided by five π plus 10. We can simplify our numerator by
noticing that negative one, always to the power of three π plus two, is just equal
to negative one all cubed all raised to the power of π multiplied by negative one
squared. We know that negative one squared
is just equal to one. Similarly, negative one cubed can
be evaluated to just be equal to negative one. Therefore, our numerator simplifies
to negative one all raised to the πth power.

To evaluate this, we recall the
alternating series test, which tells us that if a sequence ππ is equal to negative
one to the πth power multiplied by ππ and ππ isnβt always positive, decreasing
sequence with the limit as π approaches β of ππ equal to zero. Then we can conclude that the sum
to β of our sequence ππ converges. So letβs set the sequence ππ to
be equal to one divided by five π plus 10. First, letβs see if our sequence is
positive. Well we know π is always
positive. So five π plus 10 must always be
positive, and then the reciprocal of this is also positive. So our sequence ππ is always
positive.

Similarly, we can see as we
increase the value of π, what weβre doing is weβre making the denominator of our
fraction bigger and bigger, which means that as we increase π, the fraction gets
smaller and smaller. So our sequence ππ is
decreasing. Now, if we try to evaluate the
limit as π approaches β of one divided by five π plus 10, we see that as π is
getting bigger and bigger, the denominator of our fraction is getting bigger and
bigger. And so weβre getting closer and
closer to zero. So we know that the limit of ππ
as π approaches β is equal to zero. Therefore, what weβve shown by
alternating series test is the sum from π equal zero to β of negative one to the
πth power divided by five π plus 10 converges.

So in particular, what this tells
us is that negative one must be in our interval of convergence. Now, letβs substitute π₯ is equal
to one into our power series to determine whether one is in our interval of
convergence. We get the sum from π equals zero
to β of one raised to the power of three π plus two all divided by five π plus
10. One to the power of three π plus
two is just equal to one. So we get the sum from π equals
zero to β of one divided by five π plus 10. What we have is the sum of one
divided by a linear factor. This is very similar to the
harmonic series which we know diverges.

So letβs use a comparison test with
the harmonic series. In fact, letβs use the limit
comparison test, which tells us if we have two nonnegative sequences, ππ and ππ,
where the limit as π approaches β of ππ divided by ππ is positive. Then we can conclude that the sum
to β of ππ and the sum to β of ππ either both converge or both diverge. So letβs set our first sequence to
be equal to one over π and our second sequence to be equal to one divided by five
π plus 10. This gives us the limit as π
approaches β of ππ divided by ππ is equal to the limit as π approaches of one
over π divided by one over five π plus 10.

Instead of dividing by one over
five π plus 10, we can flip the fraction and multiply, giving us the limit as π
approaches β of one over π times five π plus 10 divided by one. So now, we have to calculate the
limit as π approaches β of five π plus 10 all divided by π. We can evaluate our division by π
to give us the limit as π approaches β of five plus 10 divided by π. Since π is approaching β, 10
divided by π is approaching zero. So our limit is equal to the limit
as π approaches β of the constant five, which is just equal to five. Therefore, since we have shown that
we have two nonnegative sequences and the limit as π approaches β of ππ over ππ
is positive, we can use the limit comparison test.

We know that the sum to β of the
harmonic series diverges. So we can conclude by our limit
comparison test that the sum from π equals zero to β of one divided by five π plus
10 must also diverge. What this means, in particular, is
the value of π₯ is equal to one is not in our interval of convergence. So what we have shown is that the
interval of convergence for the power series the sum from π equals zero to β of π₯
to the power of three π plus two all divided by five π plus 10 is equal to the
interval starting at negative one and including negative one going all the way up to
one, which it does not include.