Use the Direct Comparison Test to determine whether the series converges.∑n=1∞3n−2\sum_{n=1}^{\infty}\frac{3}{\sqrt{n}-2}

Diverges since a n > b n a_{n}>b_{n}

Use the Direct Comparison Test to determine whether the series converges. ∑ n = 1 ∞ 3 n − 2 \sum_{n=1}^{\infty}\frac{3}{\sqrt{n}-2}

So here, we're asked to use the direct comparison test to determine whether each of the series converges, and we'll go ahead and start with example a here. In this problem, we have the sum from n equals one to infinity of three over the square root of n minus two. So how do we go about solving a problem like this? Well, looking at this, I need to think about the tests that will apply here. And it doesn't appear that there's any kind of straightforward test I can use from the integral test, or this doesn't look like any kind of special series like a geometric or a p series specifically. So what I'm going to do is try a direct comparison here, and this is what we're told to do with the problem. It's important to understand when we need to use direct comparison test. So how could I do this? Well, what I'm gonna do is create another series over here. And I can see that the series that we have is this piece where we have this a sub n, right about there. And so because of this, I'm going to create another series with b sub n. And the b sub n that I'm going to choose is going to be something that I can figure out whether it converges or diverges. Now looking at this, I'm just going to go ahead and knock off this negative two at the end here. So that's going to give me that my b sub n is three over the square root of n. So this would be my b sub n. Now from here, what I can do is take that three, which is a constant, and pull it outside of the sum. And so we'll have the sum from n equals one to infinity, and we'll have three times, and it's going to be one over the square root of n, which is the same thing as n to the one half power. Now notice this resembles a p series. And in this p series, I can see that p is gonna be this exponent, one half. One half is clearly less than or equal to one, and I can see that, in this case, it's less than one. So and that's the rule for a p series. If p series is less than or equal to one, then we know the series is going to diverge. However, if p is greater than one, this is just the rule for the p series, and the series converges. Since I can see here that one half is less than one, that means my series is going to diverge. So we have that b sub n here diverges. But that doesn't quite answer the total problem here yet because this is just a series we're comparing. We have to figure out how this relates back to my a sub n, or really what we're doing is comparing to see which series is bigger. So we have this three over the square root of n minus two. This is my a sub n. And then we have this series right here, which is going to be three over the square root of n. This is my b sub n. So we have our a sub n, and we have our b sub n. And which of these series is bigger in the long run? Well, since this sum is going to infinity, we just want to think in general what's going to happen as we go to a long run, as we go to infinity here. And looking at this, well, starting with this one, I can see that we have the square root of n on bottom minus two. That means we're going to have a smaller overall denominator than just having the square root of n as we go farther and farther. So So because we have a smaller denominator here, I can see that the a sub n is indeed the bigger of the series. So you can see that a sub n is bigger than b sub n. But what does this tell us? Well, we figured out in this case that b sub n diverges. And notice in this case, it's the smaller series that diverged. If the smaller series diverges, then the bigger series has to diverge as well. So, therefore, we could say that our series diverges, and this would be the solution to our problem. So this is how we can use the direct comparison test to determine if a series converges or diverges. Let's move on.

Use the Direct Comparison Test to determine whether the series converges. ∑ n = 1 ∞ 3 n − 2 \sum_{n=1}^{\infty}\frac{3}{\sqrt{n}-2}

Diverges since a n > b n a_{n}>b_{n}

Converges since a n > b n a_{n}>b_{n}

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Use the Direct Comparison Test to determine whether the series converges.
$\sum_{n = 1}^{\infty} \frac{3}{\sqrt{n} - 2}$

Diverges since $a_{n} > b_{n}$

Converges since $a_{n} > b_{n}$

Partly converges and partly diverges.

Verified step by step guidance

Step 1: Recall the Direct Comparison Test. This test states that if we have two series ∑aₙ and ∑bₙ, where 0 ≤ aₙ ≤ bₙ for all n, then: (a) If ∑bₙ converges, ∑aₙ also converges. (b) If ∑aₙ diverges, ∑bₙ also diverges.

Step 2: Identify the given series ∑(3 / (√n - 2)). To apply the Direct Comparison Test, we need to find a simpler series bₙ that we can compare to aₙ = 3 / (√n - 2). The simpler series should ideally resemble the behavior of aₙ for large n.

Step 3: Analyze the denominator √n - 2. For large values of n, √n dominates over the constant -2. Thus, the term 3 / (√n - 2) behaves similarly to 3 / √n for large n. Let bₙ = 3 / √n.

Step 4: Determine the convergence or divergence of the comparison series ∑bₙ = ∑(3 / √n). This series can be rewritten as ∑(3 * n^(-1/2)). Since the exponent -1/2 is less than -1, the p-series test tells us that ∑bₙ diverges.

Step 5: Compare aₙ and bₙ. Observe that for all n ≥ 1, aₙ = 3 / (√n - 2) > bₙ = 3 / √n because subtracting 2 from √n makes the denominator smaller, increasing the value of aₙ. Since bₙ diverges and aₙ > bₙ, the Direct Comparison Test implies that the given series ∑(3 / (√n - 2)) also diverges.

5:44m

Watch next

Master Divergence Test (nth Term Test) with a bite sized video explanation from Patrick

Start learning

Convergence Tests practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice