### Video Transcript

Use the πth term divergence test to determine whether the series the sum from π equals one to infinity of three π cubed over seven π squared plus two π is divergent or whether the test is inconclusive.

Letβs begin by recalling what we actually understand by the πth term divergence test. Remember, the πth term test for divergence says that if the limit as π approaches infinity of π sub π does not exist or itβs not equal to zero, then the series the sum from π equals one to infinity of π sub π is divergent. We also know that if the limit is equal to zero, we canβt actually tell whether the series converges or diverges. And so we say the test fails.

In this question, weβre going to let π sub π be equal to three π cubed over seven π squared plus two π. And that means we need to evaluate the limit as π approaches infinity of three π cubed over seven π squared plus two π. Now, we donβt actually need to do any manipulation here. There are three rules for determining a limit of a fraction as a variable approaches infinity. These three rules involve looking at the variables in both the numerator and denominator of the fraction. Weβll begin by looking for the highest power of π in both the numerator and the denominator.

The first rule says that if the exponent of the highest term in the numerator matches the exponent of the highest term in the denominator, the limit is the fractional ratio of the coefficients of the highest terms, of the highest powers of π. The second rule says that if the numerator has the highest term or the highest power of π, then the fraction is top heavy and the limit is infinity. And if the converse is true, if the denominator has the highest term or the highest power of π, then the fraction is called bottom heavy and the limit is zero.

Now, in our question, the highest term of π on the numerator is π cubed. The highest power of π on the denominator is π squared. The numerator has the highest term. And this tells us then that the limit as π approaches infinity of three π cubed over seven π squared plus two π must be infinity. The limit does exist and itβs not equal to zero. And so we find that the series the sum from π equals one to infinity of three π cubed over seven π square plus two π must be divergent.