### Video Transcript

Consider the series the sum from π
is equal to one to β of π cubed over three π factorial. Determine whether the series
converges or diverges.

We can try using the ratio test to
determine whether the series converges or diverges. The ratio test tells us that for
some series, which is the sum from π equals one to β of π π with, πΏ being equal
to the limit as π tends to β of the absolute value of π π plus one over π π,
that firstly, if πΏ is less than one, then our series converges absolutely. Secondly, if πΏ is greater than
one, then our series diverges. And finally, if πΏ is equal to one,
then the test is inconclusive. Now, in our case, we can see that
π π is equal to π cubed over three π factorial. Letβs quickly note that this three
π factorial in the denominator is equal to three multiplied by π factorial. And so, the factorial symbol
affects only the π and not the three.

Next, we can write down what π π
plus one is. We simply change each π in π π
to π plus one. And we see that π π plus one is
equal to π plus one cubed over three π plus one factorial. Now, weβre ready to find πΏ. We have that πΏ is equal to the
limit as π tends to β of the absolute value of π π plus one over π π. We substitute in π π plus one and
one over π π. Now, weβre ready to start
simplifying.

First, we can cross cancel the
factor of three. Next, we know that we can rewrite
π plus one factorial as π plus one multiplied by π factorial. This will allow us to write our
limit as the limit as π tends to β of the absolute value of π plus one cubed
multiplied by π factorial over π plus one multiplied by π factorial multiplied by
π cubed. And so, we can cross out the factor
of π factorial from the numerator and denominator. We can also cancel a factor of π
plus one from the denominator with one of the factors of π plus one from the
numerator. What weβre left with here is the
limit as π tends to β of the absolute value of π plus one squared over π
cubed.

Next, we can distribute the square
in the numerator. Here, it is clear that our highest
power of π in the numerator is π squared and our highest power of π in the
denominator is π cubed. Since weβre taking the limit as π
tends to β and the highest power of π in the denominator is greater than the
highest power of π in the numerator, this tells us that this limit will be equal to
zero. What we have found here is that our
value of πΏ for this series is zero and zero is less than one. So when we look at our rule for the
ratio test, we see that weβve satisfied condition number one. And this tells us our solution to
the question, which is that the sum from π equals one to β of π cubed over three
π factorial converges absolutely.