Question Video: Using the Limit Comparison Test Mathematics

Video Transcript

Suppose that the sequence 𝑎 𝑛 is greater than zero and the limit as 𝑛 approaches ∞ of the square root of 𝑛 multiplied by 𝑎 𝑛 is equal to 𝑐, where 𝑐 is greater than zero. What does the limit comparison test tell us about the series the sum from 𝑛 equals one to ∞ of 𝑎 𝑛?

The question gives us some information about a sequence 𝑎 𝑛. It tells us that 𝑎 𝑛 is always positive. The limit as 𝑛 approaches ∞ of the square root of 𝑛 multiplied by 𝑎 𝑛 is equal to some 𝑐, where 𝑐 is positive. And it wants us to use the limit comparison test to tell us some information about the series the sum from 𝑛 equals one to ∞ of 𝑎 𝑛.

We recall that one version of the limit comparison test tells us that if we have a sequence 𝑎 𝑛 which is greater than or equal to zero and the sequence 𝑏 𝑛 which is greater than zero for 𝑛 is greater than or equal to one. And the limit as 𝑛 approaches ∞ of the quotient 𝑎 𝑛 divided by 𝑏 𝑛 is equal to some positive constant 𝑐. Then the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 either both converge or they both diverge.

The question wants us to use the limit comparison test to tell us the convergence or divergence of the sum from 𝑛 equals one to ∞ of our sequence 𝑎 𝑛. To use the limit comparison test, we must have all of the prerequisites are true. And the question tells us that the limit as 𝑛 approaches ∞ of the square root of 𝑛 multiplied by 𝑎 𝑛 is already equal to some positive constant 𝑐. So to find the sequence 𝑏 𝑛, we could set 𝑎 𝑛 divided by 𝑏 𝑛 to be equal to the square root of 𝑛 multiplied by 𝑎 𝑛. We can divide both sides of this equation by 𝑎 𝑛 to get one divided by 𝑏 𝑛 is equal to the square root of 𝑛. And then we can take reciprocals of both sides of this equation to give us that 𝑏 𝑛 is equal to one divided by the square root of 𝑛.

So we have the limit as 𝑛 approaches ∞ of 𝑎 𝑛 divided by 𝑏 𝑛 is equal to the limit as 𝑛 approaches ∞ of the reciprocal of 𝑏 𝑛 multiplied by 𝑎 𝑛. Then we substitute 𝑏 𝑛 is equal to one divided by the square root of 𝑛. And the reciprocal of one divided by the square root of 𝑛 is just the square root of 𝑛. This gives us the limit as 𝑛 approaches ∞ of the square root of 𝑛 multiplied by 𝑎 𝑛. And from the question, we have that this limit is equal to some positive constant 𝑐.

So we’ve shown from the limit comparison test that the limit as 𝑛 approaches ∞ of 𝑎 𝑛 divided by 𝑏 𝑛 is equal to some positive constant 𝑐. The question also tells us that the sequence 𝑎 𝑛 is positive. So the prerequisite in the limit comparison test that our sequence 𝑎 𝑛 is greater than or equal to zero is also true. The last prerequisite in the limit comparison test is we need to show that the sequence 𝑏 𝑛 is greater than zero for 𝑛 greater than or equal to one. Well, we have that the sequence 𝑏 𝑛 is equal to one divided by the square root of 𝑛. And since the square root of 𝑛 is a positive number, this is one divided by a positive number. So this is greater than zero.

Since we’ve now shown that all prerequisites of our limit comparison test are true, we must have that the sum from 𝑛 equals one to ∞ of the sequence 𝑎 𝑛 and the sum from 𝑛 equals one to ∞ of the sequence 𝑏 𝑛 both converge. Or they both diverge. So let’s consider the sum from 𝑛 equals one to ∞ of the sequence 𝑏 𝑛. Well, this is equal to the sum from 𝑛 equals one to ∞ of one divided by the square root of 𝑛. In fact, since the square root of 𝑛 is equal to 𝑛 raised to the power of a half, we have that this series is equal to the sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the power of a half, which is a 𝑝-series.

And we know that a 𝑝-series, which is the sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the 𝑝th power, will converge when 𝑝 is greater than one and will diverge when 𝑝 is less than or equal to one. So we have that our sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is equal to a 𝑝-series, where 𝑝 is equal to a half, which is less than one. So our sum from 𝑛 equals one to ∞ of 𝑏 𝑛 must diverge. And we showed using the limit comparison test that the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 either both converge or diverge.

Therefore, using the limit comparison test, since the sum from 𝑛 equals one to ∞ of one divided by the square root of 𝑛 diverges, we must have that the sum from 𝑛 equals one to ∞ of our sequence 𝑎 𝑛 is divergent.