### Video Transcript

Using the πth-term test, determine
whether the series the sum from π equals zero to β of four to the πth power
divided by three to the πth power is divergent or the test fails.

The question wants us to use the
πth-term test on the series the sum from π equals zero to β of four to the πth
power divided by three to the πth power. And we recall that the πth-term
divergence test tells us that if the limit as π approaches β of the sequence π π
is not equal to zero, then the infinite sum of π π diverges. And if the limit as π approaches β
of the sequence π π is equal to zero, then the test is inconclusive.

Since the question wants us to use
this test on the series the sum from π equals zero to β of four to the πth power
divided by three to the πth power, weβll set π π to be equal to four to the πth
power divided by three to the πth power. So letβs now calculate the limit as
π approaches β of our sequence π π. Thatβs equal to the limit as π
approaches β of four to the πth power divided by three to the πth power.

Next, weβre going to use that π₯ to
the πth power divided by π¦ to the πth power is equal to π₯ divided by π¦ all
raised to the πth power to rewrite our limit. This gives us the limit as π
approaches β of four-thirds all raised to the πth power. This gives us the limit as π
approaches β of some constant raised to the πth power. This is a geometric sequence.

And we recall that if the absolute
value of our ratio π is less than one, then the limit as π approaches β of π to
the πth power is equal to zero. And if the absolute value of our
ratio π is greater than one, then the limit as π approaches β of π to the πth
power does not exist. In our limit, we have that π is
equal to four divided by three, which is greater than one. And using what we know about
geometric sequences, we have if the absolute value of our ratio π is greater than
one, then the limit as π approaches β of π to the πth power does not exist.

So we have that the limit as π
approaches β of four-thirds to the πth power does not exist. But this limit is equal to our
limit of π π as π approaches β. So in particular, weβve shown that
the limit of π π as π approaches β is not equal to zero. Therefore, using our πth-term
divergence test, we must have that the infinite sum of π π diverges.

So in conclusion, weβve shown if we
use the πth-term divergence test on the series the sum from π equal zero to β of
four to the πth power divided by three to the πth power, then we can conclude that
this series is divergent.