Use the Limit Comparison Test to determine whether the series converges.∑n=1∞nn2+3\sum_{n=1}^{\infty}\frac{\sqrt{n}}{n^2+3}

Converges since p = 3 2 > 1 p=\frac32>1 & L > 0 L>0

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

this problem, we are asked to use the limit comparison test to determine whether each of the series converges, and we're gonna go ahead and start here with example a. How can I solve this problem? We'll notice that we have the sum from n equals one to infinity. Now we have this is the square root of n over n squared plus three. We're specifically asked to use the limit comparison test here, so how do I go about doing this? Well, my first step whenever you're doing some kind of comparison test is to first identify what your a sub n is, and That's gonna be this piece right here in my series. So now that I know my a sub n, I need to choose a b sub n. But how do I go about doing that? Well, it needs to be something that's similar to the a sub n that I have, but something that I also know I can more easily determine its convergence. And I think something that's gonna make things more simple is if I knock off this three in the denominator. So that's gonna give me the square root of n on top and then just n squared on the bottom. And this right here would be my b sub n. Now I want you to notice something interesting that happens here. I can take this and I can actually rewrite things because the square root of n over n squared, this is the same thing as n to the one half over n squared. Since I'm dividing and we have the same base here, I can just subtract the exponents, giving me n to the one half minus two. N to the one half minus two is the same thing as n to the negative three halves. And And I can just rewrite this whole thing as one over n to the three halves. So if I wanna rewrite the sum, I can see that this whole thing is equal to the sum from n equals one to infinity of one over n to the three halves power. This is the same b sub n that we have right here. Now I want you to notice something about b sub n. This is a p series now that we've rewritten this. And for a p series, p is equivalent to whatever exponent is on your end in the denominator. In this case, I can see that's three halves. Now for a p series, we have this rule. If p is less than or equal to one, the series diverges. But if it's greater than one, it converges. I clearly see three halves has a higher numerator than a denominator. That means it's going to be greater than one. So because p is greater than one, we can say that our b sub n here converges, and this is the series that we chose. Now since we figured out that b sub n converges, what I need to now do is apply the limit comparison test to what we have for a sub n. And I can do that by taking the limit as n approaches infinity for a sub n over b sub n. So in this case, we're going to have a sub n, which is the square root of n over n squared plus three. And all of this is going to be over b sub n, which I'm actually gonna go ahead and use this version over here. And I'll show you why in a moment why I'm choosing that. So this is going to be our a sub n over b sub n where we need to now find this limit as n approaches infinity. Now what I can do is take this bottom fraction here, and I can flip it and bring it to the top. So we need the reciprocal here. And what that's gonna give me is the square root of n over n squared squared plus three multiplied by n squared over the square root of n. Now this right here is why I used this version of b sub n rather than the one that we rewrote. Because notice by keeping the square root in the top here, I can cancel the square root of n completely. And that's just a rule of thumb that I learned. So since I see that we have a square root of n on top, I knew that I should probably pick the version of this that also has a square root of n. So and keep in mind, you could have used this and you would still be able to figure out the problem, but I'm just trying to make this a little easier on us. So keep this limit as it approaches infinity, then we'll have n squared over n squared plus three. Now what I need to do from here is I need to go ahead and figure out where my highest exponent is on n in the numerator and the denominator. Well, I see for the numerator, that's this n squared, and for the denominator, it's also n squared. So I need to take a ratio of the coefficients in front here, which is just going to be equal to one over one or just one. So this is what we calculated for our limit, and this is what we call l. So l here is equal to one. Now I know something for both cases when it comes to a limit comparison test. If l is greater than zero at any point, then this is going to match whatever happens to the b sub n. In this case, I can see l is clearly greater than zero. L is equal to one. And we know that b sub n converged. So because of that, we can say that our series here also converges. So that means that a sub n converges as well, and this right here would be our solution to the problem. So that is how you can use the limit comparison test. Let's move on and get some more practice.

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

Converges since p = 3 2 > 1 p=\frac32>1 & L > 0 L>0

Diverges since p = 3 2 > 1 p=\frac32>1 & L > 0 L>0

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

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

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

Diverges since $p equals three halves is greater than 1$ & $L is greater than 0$

Converges since $p equals three halves is greater than 1$ & $L is greater than 0$

Diverges since $b_{n}$ converges and $a_{b} > b_{n}$

Verified step by step guidance

Step 1: Recall the Limit Comparison Test. This test is used to determine the convergence or divergence of a series by comparing it to another series whose behavior is known. Specifically, if $ a_n $ and $ b_n $ are positive terms, and $ \lim_{n \to \infty} \frac{a_n}{b_n} = L $, where $ L $ is a finite positive number, then both series $ \sum a_n $ and $ \sum b_n $ either converge or diverge together.

Step 2: Identify the given series $ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2+3} $. To apply the Limit Comparison Test, choose a comparison series $ b_n $ that has a similar dominant term as $ a_n $. Here, the dominant term in the denominator of $ a_n $ is $ n^2 $, so we choose $ b_n = \frac{1}{n^{3/2}} $, which simplifies the comparison.

Step 3: Compute the limit $ \lim_{n \to \infty} \frac{a_n}{b_n} $. Substitute $ a_n = \frac{\sqrt{n}}{n^2+3} $ and $ b_n = \frac{1}{n^{3/2}} $ into the ratio: $ \frac{a_n}{b_n} = \frac{\frac{\sqrt{n}}{n^2+3}}{\frac{1}{n^{3/2}}} = \frac{\sqrt{n} \cdot n^{3/2}}{n^2+3} $. Simplify the expression to $ \frac{n^2}{n^2+3} $.

Step 4: Evaluate $ \lim_{n \to \infty} \frac{n^2}{n^2+3} $. As $ n \to \infty $, the term $ +3 $ in the denominator becomes negligible compared to $ n^2 $. Thus, $ \lim_{n \to \infty} \frac{n^2}{n^2+3} = 1 $. Since $ L = 1 $ is a finite positive number, the Limit Comparison Test applies.

Step 5: Determine the behavior of $ \sum b_n = \sum \frac{1}{n^{3/2}} $. This is a p-series with $ p = \frac{3}{2} > 1 $, which converges. By the Limit Comparison Test, since $ \sum b_n $ converges and $ L > 0 $, the given series $ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2+3} $ also converges.

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Use the Limit Comparison Test to determine whether the series converges.∑n=1∞nn2+3\(\sum\)_{n=1}^{\(\infty\)}\(\frac{\sqrt{n}\)}{n^2+3}

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