9–36. Comparison tests Use the Comparison Test or the Limit Co...

Question

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.

∑ (k = 1 to ∞) sin(1 / k) / k²

Accepted Answer

Use the direct comparison test or limit comparison test to determine the convergence of the sum from 1 to infinity of sin inverse of 1 divided by M, all divided by M squared. We then want to see if it converges or diverges. Now, let's look at our general term. We'll take a sub M equals the sign inverse of one divided by M, all divided by Msquared. Now, we have the small angle approximation for arcs sign, which tells us sine inverse. Of X is approximately X as X approaches 0. So then we can say the sign in verse of 1 divided by M approaches approximately 1 divided by M. So we will then say A subm is approximately 1 divided by m. Divided by M square. Which just gives us 1 divided by M to the 3rd. Now let's choose our comparison series. We will say B M equals 1 divided by M to the 3rd. Now by P series, This is Convergent. Because the P is the X1 and 3, which is greater than 1. So now, we can apply the limit comparison tips. This tells us that if we have a limit, As M goes to infinity of a sub m. Divided by B M This limit should be between 0 and infinity. If this is true, Both But A of And be of him sums. Either converge or diverge. And they will converge or diverge together. So now, we can apply our limit comparison. So, we have L. Equals the limit as m approaches infinity. Of our series. Which in this case, this is sign and verse. Of one divided by m, all divided by m squared. Which is all divided by 1, divided by M to the 3rd. We can simplify this. To get the limit, as M approaches infinity. A sign in verse. One divided by M, all divided by 1 divided by M. Now, let's let X equals 1 divided by M. We have the limit as X right now. Approaches 0 Of sine in verse x divided by x. Now, this is A limit property that we know, and this just equals one. In other words, L equals 1. We have confirmed that 1 is between 0 and infinity, and because BM converges, A sub-M also converges by the limit test. This means the answer to our problem is answer A. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.

∑ (k = 1 to ∞) sin(1 / k) / k²

Key Concepts

Comparison Test

Limit Comparison Test

Behavior of sin(1/k) for large k

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