# Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution Mathematics • Higher Education

Use the comparison test to determine whether the series ∑_(𝑛 = 1)^(∞) ln 𝑛/𝑛 is convergent or divergent.

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### Video Transcript

Use the comparison test to determine whether the series which is the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 over 𝑛 is convergent or divergent.

We can start answering this question by recalling what the comparison test tells us. The comparison test tells us that for the series which is the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, where 𝑎 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. That if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is convergent, and 𝑎 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 is also a convergent. And if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is divergent, and 𝑎 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 is also divergent.

Now, the series we’ve been given in the question is the sum from 𝑛 equals one to ∞ of the natural algorithm of 𝑛 over 𝑛. Therefore, we can let 𝑎 𝑛 be equal to the natural logarithm of 𝑛 over 𝑛. Now, we need to find another series to compare this to. Looking at 𝑎 𝑛, we can see that it looks similar to a 𝑝 series. Now, a 𝑝 series is a series of the form of the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝. And this 𝑝 series will diverge if 𝑝 is less than or equal to one and converge if 𝑝 is greater than one.

Now, the 𝑝 series, which our series looks most similar to, is the sum from 𝑛 equals one to ∞ of one over 𝑛. For this series, the value of 𝑝 is one. And therefore, the series will diverge. Before we can attempt to use the comparison test, we first need to check whether 𝑎 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all 𝑛. Now, the values of 𝑛 which we’ll be using go from one to ∞. Therefore, 𝑛 is greater than or equal to zero.

Now, the natural logarithm of 𝑛 is an increasing function. And when 𝑛 is equal to one, the natural logarithm of 𝑛 is equal to zero. Therefore, we can say that for all our values of 𝑛, the natural logarithm of 𝑛 will be greater than or equal to zero. Since these two things are both greater than or equal to zero, this means that both 𝑎 𝑛 and 𝑏 𝑛 will be greater than or equal to zero too.

Now, we’ve already said that the sum from 𝑛 equals one to ∞ of one over 𝑛 is divergent. Therefore, we need to try and satisfy the second part of the comparison test. Which means that we need to show that 𝑎 𝑛 is greater than or equal to 𝑏 𝑛. In other words, that’s showing that the natural logarithm of 𝑛 over 𝑛 is greater than or equal to one over 𝑛. Since 𝑛 is always greater than or equal to one, we can multiply both sides by 𝑛. Therefore, we have that the natural logarithm of 𝑛 is greater than or equal to one. And we need to try and prove this for all 𝑛.

Now, we can immediately see that this cannot be the case since the natural logarithm of one is equal to zero, which is less than one. However, this doesn’t mean that we cannot use the comparison test. Let’s consider the first few values of the natural logarithm of 𝑛. We have that the natural logarithm of two is equal to 0.693 and so on. And the natural logarithm of three is equal to 1.098 and so on.

Therefore, the natural logarithm of three is in fact greater than or equal to one. And as we said earlier, we know that the natural logarithm of 𝑛 is an increasing function. And so, our statement that the natural logarithm of 𝑛 is greater than or equal to one holds true for values of 𝑛 which are greater than or equal to three.

Now, we can rewrite our series which we were given in the question. We can take the first two terms out of the series. And we have that the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 over 𝑛 is equal to the natural logarithm of one over one plus the natural logarithm of two over two plus the sum from 𝑛 equals three to ∞ of the natural logarithm of 𝑛 over 𝑛. Now, we can compare this series with the 𝑝 series of the sum from 𝑛 equals three to ∞ of one over 𝑛.

We have the same values of 𝑎 𝑛 and 𝑏 𝑛 from before. And again, we’ve of course satisfied the condition that 𝑎 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. And we have that the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is divergent. And we also now have that 𝑎 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛. And so, from this, we can say that the sum from 𝑛 equals three to ∞ of the natural logarithm of 𝑛 over 𝑛 is divergent.

We can now relate this back to the original sum we were given in the question since its equal to two constant terms plus the series we’ve just shown to be divergent. Since it’s equal to the sum of two constants and a divergent series, we have found that the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 over 𝑛 is divergent.