### Video Transcript

Using the ratio test, determine
whether the series the sum from π equals one to β of negative two to the πth power
times π to the fourth power all over π factorial is convergent, divergent, or
whether the test is inconclusive.

Letβs begin by recalling what we
mean by the ratio test. Given a series defined as the sum
of π π, letβs let πΏ be equal to the limit as π approaches β of the absolute
value of π sub π plus one over π sub π. If πΏ, the constant, is less than
one, then that tells us that the series is absolutely convergent and hence
convergent. If πΏ is greater than one, the
series diverges. And if πΏ is equal to one, the test
is inconclusive. In this case, weβre going to define
π sub π to be equal to negative two to the πth power times π to the fourth power
all over π factorial.

To generate π sub π plus one, we
replace π with π plus one. So we get negative two to the power
of π plus one times π plus one to the fourth power over π plus one factorial. πΏ is the limit as π approaches β
of the absolute value of the quotient of these. So thatβs the absolute value of
negative two to the power of π plus one times π plus one to the fourth power over
π plus one factorial divided by negative two to the πth power times π to the
fourth power over π factorial.

Next, we recall that to divide by a
fraction, we multiply by the reciprocal of that fraction. So we had the limit as π
approaches β of the absolute value of negative two to the power of π plus one times
π plus one to the fourth power over π plus one factorial times π factorial over
negative two to the πth power times π to the fourth power. We then recall that to divide two
numbers whose bases are equal, we simply subtract the exponents. So negative two to the power of π
plus one divided by negative two to the πth power is negative two to the power of
π plus one minus π. Well, thatβs negative two to the
power of one or simply negative two. So we can cross cancel this
negative two to the power of π.

Similarly, we know π plus one
factorial is equal to π plus one times π times π minus one and so on. Or itβs equal to π plus one times
π factorial. And we can see that we can divide
through by a factor of π factorial. And then the bit inside the limit
simplifies quite nicely. Itβs the absolute value of negative
two times π plus one to the fourth power over π plus one times π to the fourth
power. We can see we can divide through by
a factor of π plus one. And we can also take out this
constant factor of negative two. Remembering, of course, it needs to
be the absolute value of negative two, which is simply two. So we have two times the limit as
π approaches β of π plus one cubed over π to the fourth power.

Now, a nice way to see what happens
on the next step is to distribute the parentheses. π plus one cubed is π cubed plus
three π squared plus three π plus one. And we could achieve this simply by
distributing the parentheses in turn or using the binomial theorem. And now, we look to manipulate this
fraction somewhat. We canβt just substitute in π
equals β. If we do, we get β over β, which
is, of course, indeterminant. So, instead, we divide the
numerator and the denominator of our fraction by the highest power of π in the
denominator, by π to the fourth power. When we do, the denominator becomes
one. And our numerator becomes π cubed
over π to the fourth power, which is one over π, plus three π squared over π to
the fourth power, which is three over π squared. Plus three π over π to the fourth
power, which is three over π cubed, plus one over π to the fourth power.

Now, as π₯ grows larger, one over
π grows smaller, three over π squared grows smaller, three over π cubed grows
smaller, and one over π to the fourth power grows smaller. In other words, as π approaches β,
each of these individual fractions approaches zero. And so, we have two times the
absolute value of zero, which is simply zero. And weβve found the value of
πΏ. Now, of course, zero is less than
one. So weβre going to use the first
part of the ratio test. And we say that the series
converges.