Use the Alternating Series Test to determine the convergence or divergence of the series. ∑ n = 1 ∞ ( − 1 ) n + 1 n + 2 \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{\sqrt{n}+2}

this problem, we are asked to use the alternating series test to determine the convergence or divergence of the series. So we have this sum here from n equals one to infinity, and this is negative one to the power of n plus one over the square root of n plus two. Now this is a case where I can see this is the piece that makes my series alternate, negative one to the n plus one power. So because of that, what I'm gonna do is set my a sub n equal to everything else that we have. So it's gonna be one over the square root of n plus two. I don't include this piece that makes it alternate here. I'm just going to include everything else. So we have our a sub n right here. And now that we have this as our piece here, well, what I need to do is use the the alternating series test to determine if this is convergent or divergent. Now something I'll mention here for this specific problem, we actually don't need to worry about whether the series converges conditionally or absolutely. We just need to focus on whether or not it passes the alternating series test. And for that, there are two conditions that need to be met. First off, the limit as n approaches infinity for a series a sub n has to go to zero. Secondly, we also need to have a case where a sub n is going to be decreasing for all sufficiently large values of n. So let's start by checking the first case right here. So that's going to be the limit as n approaches infinity for a sub n. Now looking at a sub n here, I can see that the only n we have is in the denominator, and this is going to infinity. Now, yes, it is underneath inside of a square root here, but this is still the only piece that's going to be growing. So as n approaches infinity, eventually, this is going to go to infinity setting this whole fraction to zero. So So because of this our limit here does come out to zero, which means our first condition for the alternating series test is met. But now we need to check the next condition. Is a sub n decreasing? Well-to-do that I'm going to check a sub n plus one, the next step up in our series. And for that we're going to have one over the square root of n plus one, replacing that n with n plus one, plus two. So this is what we get. Now is this going to be decreasing? We'll notice here that we end up with actually a bigger overall denominator than we had right there. So comparing a sub n to a sub n plus one, a bigger denominator means a smaller overall fraction. So that means that a sub n plus one is less than a sub n. So because of that, we can say, yes. This is going to be decreasing, so we can check off this next criteria as well. So because both of these criteria are met, we can say our series here indeed does converge, and that would be the solution to part a of this problem. So now that we found the solution, let's move on to part b. See you in the next one.

Use the Alternating Series Test to determine the convergence or divergence of the series. ∑ n = 1 ∞ ( − 1 ) n + 1 n + 2 \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{\sqrt{n}+2}

Cannot be determined since a n a_{n} is not decreasing.

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

Use the Alternating Series Test to determine the convergence or divergence of the series.
$\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{\sqrt{n} + 2}$

Converges

Diverges

Cannot be determined since $a_{n}$ is not decreasing.

Verified step by step guidance

Step 1: Recall the Alternating Series Test. This test states that an alternating series ∑(-1)^(n+1)a_n converges if two conditions are met: (1) The terms a_n are positive, and (2) The sequence {a_n} is monotonically decreasing and approaches 0 as n → ∞.

Step 2: Identify the sequence a_n from the given series. In this case, a_n = 1 / (√n + 2). Note that the alternating factor (-1)^(n+1) is separate from a_n.

Step 3: Check if a_n is positive. Since √n + 2 is always positive for n ≥ 1, a_n = 1 / (√n + 2) is positive for all n.

Step 4: Verify if a_n is monotonically decreasing. To do this, compare a_n and a_(n+1). Observe that as n increases, √n + 2 increases, which makes 1 / (√n + 2) decrease. Thus, a_n is monotonically decreasing.

Step 5: Confirm that lim (n → ∞) a_n = 0. As n approaches infinity, √n + 2 grows indefinitely, causing 1 / (√n + 2) to approach 0. Therefore, the sequence {a_n} satisfies all conditions of the Alternating Series Test, and the series converges.