### Video Transcript

What can we conclude by applying
the πth term divergence test in the series the sum of two times the natural log of
π over three π for π equals one to infinity?

We begin by recalling that the πth
term test for divergence says that if the limit as π approaches infinity of ππ is
not equal to zero or does not exist, then the series the sum of ππ from π equals
one to infinity is divergent. And indeed, if that limit is equal
to zero, we canβt tell whether the series converges or diverges and we say that the
test fails.

In our question then, weβre going
to let ππ be equal to two times the natural log of π over three π. And so our job is to evaluate the
limit as π approaches infinity of this expression. If we were to simply apply direct
substitution, then weβd find that our limit is equal to infinity over infinity. And of course, thatβs
indeterminate.

So instead, weβre going to recall
LβHΓ΄pitalβs rule. This says if the limit as π₯
approaches π of π of π₯ over π of π₯ is equal to infinity over infinity, then the
limit as π₯ approaches π of π prime of π₯ over π prime of π₯ will tell us the value of
the limit as π₯ approaches π of π of π₯ over π of π₯. We can also use this formula if our
limit is equal to zero over zero. But weβre not interested in that
case.

Now of course, weβre working with
π. So weβre going to need to
differentiate two times the natural log of π and three π with respect to π. The derivative of the natural log
of π is one over π. So when we differentiate two times
the natural log of π with respect π, we get two over π. And then the derivative of three π
is simply three. So we can now evaluate this as π
approaches infinity.

As π gets larger, two over π gets
smaller. And as π approaches infinity
therefore, two over π approaches zero. We find that this is therefore
equal to zero over three, which is zero. And we find that the test fails or
itβs inconclusive.