Use the Limit Comparison Test to determine if the following series converges.∑n=1∞12n2−ncos(nπ)\sum_{n=1}^{\infty}\frac{1}{2n^2-n\cos\left(n\pi\right)}

Converges because L = 1 > 0 L=1>0 and for b n b_{n} , p = 2 > 1 p=2>1

Use the Limit Comparison Test to determine if the following series converges. ∑ n = 1 ∞ 1 2 n 2 − n cos ( n π ) \sum_{n=1}^{\infty}\frac{1}{2n^2-n\cos\left(n\pi\right)}

we want to use the limit comparison test to determine if the following series converges. Now the series that we have is the sum from n equals one to infinity of one over two n squared minus n times the cosine of n pi. Now, since we're asked specifically to use the limit comparison test, well, what I'm gonna start by doing is identifying what my a sub n is. This should always be our first step when it comes to using a type of comparison test. So you can see this is my a sub n. And looking at this, I need to choose another series, b sub n. And for my next series here, for figuring out this b sub n, it's gonna be something we can compare to our a sub n. But how could I make this look a bit more simple? Well, looking at this, I see that we have this one in the numerator, and that's fine. But this whole thing in the denominator looks a bit complicated. So what I'm gonna try doing is knocking off this n times the cosine of n pi here, and and the negative sign as well. So that's just going to give me that my b sub n is one over two n squared. Now it looks like this is something that I could find the convergence of. So what I'm going to do is take this onetwo, and I'm going to pull it outside of the sum. So that'll be one half times the sum from n equals one to infinity of one over n squared. Now notice it's pretty clear that we have a p series that shows up here, because this exponent here, this is our p, and it's equal to two. And remember, whenever we have a case where p is less than or equal to one, that means that the series diverges. However, if it's greater than one, the series converges. And I can see two is clearly greater than one, so that means our b sub n is going to converge. And keep in mind, our b sub n was this series that we chose right here. I'm not gonna box this one because I know we pulled the one half out, so I'm just gonna keep this one over two n squared as my b sub n. So this here converges. But this is not the solution to the problem. This only tells us about b sub n. We need to now figure out if a sub n is going to converge, and we can do that by using the limit comparison test. So I'm gonna take the limit as n approaches infinity. Now what we're gonna have is a sub n over b sub n. My a sub n is one over this whole denominator, two n squared minus n times the cosine of n pi, and then in the here, we're going to divide this by one over two n squared, which is my b sub n. I can go ahead and flip this one over two n squared and bring it to the top here, because that's going to give me the limit as n approaches infinity, and it's going to be two n squared over, it's going to be two n squared minus n cosine of n pi. Now from here, I need to take my limit. Now to take the limit here, what I want to look for is the highest degree that we have on n. Now I can see that that's two in the numerator, and it's also two in the denominator. So because we have these n squared, I'm just going to take a ratio of the coefficients. So doing this is going to give me that this limit is two over two, which I know is just equal to one. Now recall that these are the rules when it comes to dealing with the limit comparison test. If l is greater than zero, well then what that's going to mean, this is just going to match whatever b sub n is. Now I want to remind you of a couple other rules that may occur here too, because if l was equal to zero specifically, well then, this is only going to work if this matches a case where the series converges. If If it diverges though, if b sub n were to diverge, then we couldn't actually use this rule. This is only a case for a converging series. Now likewise, if l were to explode to infinity, well, this would mean that our series diverges, but this is only the case if b sub n also diverges. Since we have a converging case, we don't need to worry about this scenario. Now what I can see here is clearly l is equal to one, which is a value that is greater than or equal to zero. So because this condition is met, we would say that our whole series here converges. So this here would be the solution to our problem. So since b sub n converges, we can say that our a sub n here converges as well. So that is how you can solve this problem to figure out if your series converges using the limit comparison test. Hope you found this helpful, and let's move on.

Use the Limit Comparison Test to determine if the following series converges. ∑ n = 1 ∞ 1 2 n 2 − n cos ⁡ ( n π ) \sum_{n=1}^{\infty}\frac{1}{2n^2-n\cos\left(n\pi\right)}

Converges because L = 1 > 0 L=1>0 and for b n b_{n} , p = 2 > 1 p=2>1

Diverges because L = 1 > 0 L=1>0 L = 1 > 0 and for b n b_{n} b n , p = 2 > 1 p=2>1 p = 2 > 1

Converges because L = 0 L=0 and for b n b_{n} b n , p = 2 > 1 p=2>1 p = 2 > 1

Diverges because L = 0 L=0 and for b n b_{n} b n , p = − 2 < 1 p=-2<1

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 if the following series converges.
$\sum_{n = 1}^{\infty} \frac{1}{2 n^{2} - n \cos (n π)}$

Converges because $L equals 1 is greater than 0$ and for $b_{n}$ , $p equals 2 is greater than 1$

Diverges because $L=1>0$ and for $b_{n}$ , $p=2>1$

Converges because $L equals 0$ and for $b_{n}$ , $p=2>1$

Diverges because $L equals 0$ and for $b_{n}$ , $p = - 2 < 1$

Verified step by step guidance

Step 1: Recall the Limit Comparison Test. This test states that for two series ∑a_n and ∑b_n with positive terms, if lim (n → ∞) (a_n / b_n) = L, where L is a positive finite number, then both series either converge or diverge together.

Step 2: Identify the given series a_n = 1 / (2n^2 - n cos(nπ)). Notice that the dominant term in the denominator as n becomes large is 2n^2. This suggests that we compare the given series with b_n = 1 / (2n^2), which simplifies to b_n = 1 / n^2.

Step 3: Compute the limit of the ratio a_n / b_n as n approaches infinity. Substitute a_n and b_n into the ratio: (a_n / b_n) = (1 / (2n^2 - n cos(nπ))) / (1 / n^2). Simplify this expression to get (a_n / b_n) = n^2 / (2n^2 - n cos(nπ)).

Step 4: Evaluate the limit lim (n → ∞) (n^2 / (2n^2 - n cos(nπ))). Factor n^2 out of the denominator: lim (n → ∞) (n^2 / (n^2(2 - cos(nπ) / n))). Simplify further to get lim (n → ∞) (1 / (2 - cos(nπ) / n)). As n → ∞, cos(nπ) / n → 0, so the limit becomes 1 / 2.

Step 5: Since the limit L = 1/2 > 0 and b_n = 1 / n^2 is a p-series with p = 2 > 1 (which converges), the given series ∑a_n also converges by the Limit Comparison Test.

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Use the Limit Comparison Test to determine if the following series converges.∑n=1∞12n2−ncos(nπ)\(\sum\)_{n=1}^{\(\infty\)}\(\frac{1}{2n^2-n\cos\left(n\pi\right)}\)

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Use the Limit Comparison Test to determine if the following series converges.
$\sum_{n = 1}^{\infty} \frac{1}{2 n^{2} - n \cos (n π)}$