45–63. Absolute and conditional convergence Determine whether ...

Question

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (k = 1 to ∞) (−1)ᵏ · k / (2k + 1)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine whether the series, the sum of K equals 1 to infinity of the quantity of -1 in quantity ratio power K multiplied by K, divided by the quantity of 3K plus 2 in quantity, converges absolutely, converges conditionally, or diverges. So we're going to begin by looking at whether the series converges absolutely. So for that, we need to look at the sum of K equal to 1 to infinity of the absolute value of the quantity of the quantity of minus 1 in quantity rates of the power k multiplied by k divided by the quantity of 3K plus 2 in quantity. And we can actually simplify this expression by taking the absolute value. So this is just going to be equal to the sum of K equal to 1 to infinity of K divided by the quantity of 3K plus 2 in quantity. So to see whether this series converges or diverges, we're going to use the divergence test. So for that we need to look at the limit as K approaches infinity of K divided by the quantity of 3K plus 2 in quantity, and this limit is equal is not equal to 0, then that means that our series here diverges. Now we can rewrite the fraction here by dividing both the numerator and denominator by K. So this is going to be equal to the limit as K approaches infinity of 1 divided by the quantity of 3 plus 2 divided by k in quantity. And as K approaches infinity, 2 divided by k approaches 0, so that means that this limit is just going to be equal to 1 divided by 3, which is not equal to 0. So that means that our series here diverges. So that means that our original series does not converge absolutely. So next we need to see whether our original series converges or not, because if it does, then it means that our series converges conditionally. So here we're gonna have to look at the sum. Of K equal to 1 to infinity of the quantity of minus 1 in quantity raised to the power K multiplied by K divided by the quantity of 3K plus 2 in quantity. Which we can rewrite as just being equal to the sum of K equal to 1. To infinity of the quantity of -1 and quantity rates to the power K. Multiplied by the quantity of k divided by the quantity of 3k plus 2 in quantity. So that means our series is of the form of the sum of k equal to 1 to infinity of the quantity of minus 1 in quantity raises of power k multiplied by a sub k, where a sub k is equal to k divided by the quantity of 3k plus 2 in quantity. So we have an alternating series here and so we're going to use the alternating series test. And so we call for a series to converge using an alternating series test, two conditions must be met. That the limit as k approaches infinity of a sub K has to be equal to 0, and a K plus 1 has to be less than a subk since A subk needs to be a decreasing quantity as k increases. Now, if we look at the limit. As K approaches infinity of A to K, we'll have the limit as K approaches infinity of K divided by the quantity of 3K plus 2 in quantity. And we already found that this limit is equal to 1 divided by 3. So, since this is not equal to 0, that means that this series is also going to diverge. And so that is going to be the answer that we're looking for, for this problem. So our series here diverges. So, I hope that this explanation has been useful, and I'll see you in the next video.

45–63. Absolute and conditional convergence Determine whether the following series converge absolutely, converge conditionally, or diverge.
∑ (k = 1 to ∞) (−1)ᵏ · k / (2k + 1)

Key Concepts

Absolute Convergence

Conditional Convergence

Alternating Series Test

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