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

What can we conclude by applying the 𝑛th term divergence test in the series βˆ‘_(𝑛 = 1) ^(∞) (2ln 𝑛/3𝑛)?

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Video Transcript

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

We begin by recalling that the 𝑛th term test for divergence says that if the limit as 𝑛 approaches infinity of π‘Žπ‘› is not equal to zero or does not exist, then the series the sum of π‘Žπ‘› from 𝑛 equals one to infinity is divergent. And indeed, if that limit is equal to zero, we can’t tell whether the series converges or diverges and we say that the test fails.

In our question then, we’re going to let π‘Žπ‘› be equal to two times the natural log of 𝑛 over three 𝑛. And so our job is to evaluate the limit as 𝑛 approaches infinity of this expression. If we were to simply apply direct substitution, then we’d find that our limit is equal to infinity over infinity. And of course, that’s indeterminate.

So instead, we’re going to recall L’HΓ΄pital’s rule. This says if the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to infinity over infinity, then the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯ will tell us the value of the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯. We can also use this formula if our limit is equal to zero over zero. But we’re not interested in that case.

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

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

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