Question Video: Making a Conclusion by Applying the 𝑛th-Term Divergence Test to a Series | Nagwa Question Video: Making a Conclusion by Applying the 𝑛th-Term Divergence Test to a Series | Nagwa

# Question Video: Making a Conclusion by Applying the πth-Term Divergence Test to a Series Mathematics • Higher Education

What can we conclude by applying the πth term divergence test in the series β_(π = 1) ^(β) (2ln π/3π)?

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### Video Transcript

What can we conclude by applying the πth 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 πth term test for divergence says that if the limit as π approaches infinity of ππ is not equal to zero or does not exist, then the series the sum of ππ from π equals one to infinity is divergent. And indeed, if that limit is equal to zero, we canβt tell whether the series converges or diverges and we say that the test fails.

In our question then, weβre going to let ππ 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 weβd find that our limit is equal to infinity over infinity. And of course, thatβs indeterminate.

So instead, weβre going to recall LβHΓ΄pitalβs 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 weβre not interested in that case.

Now of course, weβre working with π. So weβre 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 itβs inconclusive.

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