Use the Direct Comparison Test to determine whether the series converges.∑n=1∞cos2nn2\sum_{n=1}^{\infty}\frac{\cos^2n}{n^2} Hint: Compare to bn=1n2b_{n}=\frac{1}{n^2}

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

Use the Direct Comparison Test to determine whether the series converges. ∑ n = 1 ∞ cos 2 n n 2 \sum_{n=1}^{\infty}\frac{\cos^2n}{n^2} Hint: Compare to b n = 1 n 2 b_{n}=\frac{1}{n^2}

Use the direct comparison test to determine whether each of the series converges. Now in the last video, we did example a, so now we'll move to part b or example b. So here we have the sum from n equals one to infinity, and we have cosine squared n over n squared. And we're given a hint here to use the b sub n as one over n squared. So how do I solve this problem? Well, this is a case where we have this series, and I can see that there's not really a straightforward test that we already know about that we can use. This doesn't resemble any kind of special series, like a p series, harmonic, or geometric series. And I can also tell just by looking at this that I'm not gonna be able to do an integral test very easily with this cosine squared on top. So we need to use a comparison test, and we're actually given what we're comparing this to. So in this problem, we have our a sub n, which is given to us. That's gonna be the series we have in question, and we're going to choose our series b sub n. So we'll go from n equals one to infinity of b sub n, which we're told is going to be one over n squared. So notice here for our series, we're just going ahead and knocking off this cosine squared n on top and replacing that with just this one here. And what I can do is recognize something about b sub n. Notice that this is our b sub n. This resembles a p series. Because, see, we have one over n to an exponent. The exponent we're dealing with is two. Now we know for a p series, if p is less than one or less than or equal to one, the series is going to diverge. However, if p is greater than one, then the series is going to converge. I can see right here that p is two, which is clearly greater than one. So because we have that our p here is greater than one, well, that means that our series is going to converge. So we can say b sub n here converges. But remember, this doesn't necessarily tell me the solution to my problem, because all we did was figure out that our b sub n converged. But how does this relate to our a sub n? Well, what I need to do is take a look at the two series that I have. So this is the series we have in question, cosine squared n over n squared, and then we're comparing this to b sub n, which is one over n squared. So this is our a sub n, and that is our b sub n. Now this is where things can get a bit tricky with this problem. You need to determine which one is the greater series as we go to the long run here. Now it may be tempting to look at this cosine squared n and think, well, since we have an n on top here, that means that this is going to grow faster, and so this this therefore, it's going to be the bigger series in the long run. But that's actually not necessarily true, because think about how the cosine behaves. And I'll go ahead and draw just the graph over here. When we are dealing with the cosine that we're graphing, the cosine goes between one and negative one and oscillates and looks something kind of like this, and it just keeps on going. So we go back and forth between one and negative one, so that's going to be what happens as we go from one to infinity. So we'll go with these integer values one, two, three, and four. And what we're what's gonna happen is at most, we're gonna get a one for an output as we go up, and at least we'll get negative one. But notice that we have cosine squared. So because of that, we're not going to get a negative value. So all of our values that we'll get for the output of cosine squared n here, they're going to basically end up just being between zero and one. So these are the values that we'll get as an output for our cosine squared n. Now because those are the values that we get as an output, notice here that this denominator here is going to keep increasing. But notice this is just gonna be between zero and one as we go up. So, actually, in the long run, it's going to be this series b sub n, which is greater than the series cosine squared n over n squared. So we can actually see b sub n is the greater series since at most will be at one here. But in this case, in the long run, we can't guarantee that this one is going to be ahead of this series. In fact, it's actually going to be smaller, so b sub n is the bigger series. So notice that we had a case where the bigger series b sub n converged. And because of that, that verifies that our entire series here for a sub n also converges. Because if the bigger series converges, then the smaller series has to converge as well. So this would be our solution to the problem. Hope you found this helpful.

Use the Direct Comparison Test to determine whether the series converges. ∑ n = 1 ∞ cos ⁡ 2 n n 2 \sum_{n=1}^{\infty}\frac{\cos^2n}{n^2} Hint: Compare to b n = 1 n 2 b_{n}=\frac{1}{n^2}

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

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

Diverges since b n b_{n} converges

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

Use the Direct Comparison Test to determine whether the series converges.
$\sum_{n = 1}^{\infty} \frac{\cos^{2} n}{n^{2}}$ Hint: Compare to $b_{n} = \frac{1}{n^{2}}$

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

Diverges since $b_{n}$ converges

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

Verified step by step guidance

Step 1: Recall the Direct Comparison Test. This test states that if 0 ≤ a_n ≤ b_n for all n and the series ∑b_n converges, then the series ∑a_n also converges. Similarly, if ∑b_n diverges and a_n ≥ b_n for all n, then ∑a_n diverges.

Step 2: Identify the given series and the comparison series. The given series is ∑(cos²n / n²), and the comparison series is b_n = 1 / n². Note that b_n is a p-series with p = 2, which is known to converge because p > 1.

Step 3: Analyze the term a_n = cos²n / n². Since cos²n is always between 0 and 1 (because the cosine function oscillates between -1 and 1, and squaring it ensures non-negative values), we have 0 ≤ cos²n ≤ 1. This implies 0 ≤ a_n ≤ b_n for all n.

Step 4: Apply the Direct Comparison Test. Since 0 ≤ a_n ≤ b_n and the series ∑b_n converges, the Direct Comparison Test guarantees that the series ∑a_n also converges.

Step 5: Conclude that the given series ∑(cos²n / n²) converges by the Direct Comparison Test.

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