# Question Video: Determining If a Given Series Is Divergent Using the πth Term Test Mathematics • Higher Education

Using the πth-term test, determine whether the series β_(π = 0)^(β) 4^(π)/3^(π) is divergent or the test fails.

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### Video Transcript

Using the πth-term test, determine whether the series the sum from π equals zero to β of four to the πth power divided by three to the πth power is divergent or the test fails.

The question wants us to use the πth-term test on the series the sum from π equals zero to β of four to the πth power divided by three to the πth power. And we recall that the πth-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 πth power divided by three to the πth power, weβll set π π to be equal to four to the πth power divided by three to the πth power. So letβs now calculate the limit as π approaches β of our sequence π π. Thatβs equal to the limit as π approaches β of four to the πth power divided by three to the πth power.

Next, weβre going to use that π₯ to the πth power divided by π¦ to the πth power is equal to π₯ divided by π¦ all raised to the πth power to rewrite our limit. This gives us the limit as π approaches β of four-thirds all raised to the πth power. This gives us the limit as π approaches β of some constant raised to the πth 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 πth 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 πth 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 πth power does not exist.

So we have that the limit as π approaches β of four-thirds to the πth power does not exist. But this limit is equal to our limit of π π as π approaches β. So in particular, weβve shown that the limit of π π as π approaches β is not equal to zero. Therefore, using our πth-term divergence test, we must have that the infinite sum of π π diverges.

So in conclusion, weβve shown if we use the πth-term divergence test on the series the sum from π equal zero to β of four to the πth power divided by three to the πth power, then we can conclude that this series is divergent.