# Question Video: The πth Term Divergence Test Mathematics • Higher Education

Use the πth term divergence test to determine whether the series β_(π = 1) ^(β) (πΒ³ β πβ΅)/(6πβ΅ + πΒ²) is divergent or whether the test is inconclusive.

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### Video Transcript

Use the πth term divergence test to determine whether the series the sum from π equals one to infinity of π cubed minus π to the fifth power over six π to the fifth power plus π squared is divergent or whether the test is inconclusive.

And letβs begin by recalling what we understand about the πth term divergence test. It tells us that if the limit as π approaches infinity of π sub π does not exist or is not equal to zero, then the series the sum from π equals one to infinity of π sub π is divergent. And so we define π sub π to be equal to π cubed minus π to the fifth power over six π to the fifth power plus π squared. And this means we need to evaluate the limit as π approaches infinity of π cubed minus π to the fifth power over six π to the fifth power plus π squared.

We canβt do this by direct substitution. So instead, we perform a little trick. We divide both the numerator and the denominator of our fraction by the highest power of π in the denominator, so by π to the fifth power. Letβs do that term by term. The first term on our numerator is π cubed. So we have π cubed over π to the fifth power, which is equal to one over π squared. We then subtract π to the fifth power over π to the fifth power, which is one. Then on the denominator, we have six π to the fifth power over π to the fifth power, which is six, plus π squared over π to the fifth power, which is one over π cubed.

And this step is really useful because we now have two constants, negative one and six. But also as π grows larger, one over π squared and one over π cubed grow smaller. In fact, as π approaches infinity, these two terms approach zero. So the limit as π approaches infinity of one over π squared minus one over six plus one over π cubed is negative one-sixth. This is not equal to zero. And by the πth term divergence test, that tells us that the series is divergent. And so we say that the series is divergent. And remember, of course, that if the limit had been equal to zero, we canβt actually tell whether this series converges or diverges. And so we say the test fails or itβs inconclusive.