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 ∞) ((3k³ + 4)(7k² + 1)) / ((2k³ + 1)(4k³ − 1))

Accepted Answer

Use the direct comparison test or limit comparison test to determine the convergence of the sum from 1 to infinity of 45 plus 1 multiplied by 23 plus 3, all divided by 3M 5th plus 2 multiplied by +55 minus 4. And we're seeing if it either converges or diverges. Now, let's first find our two comparison sums. In this case, we will make use of the limit comparison tests. And we will first determine a sub M. Now we'll use our current series. We'll say 4M to the 5th, multiplied by 2 M to the 3rd. Divided by 3m to the 5th, multiplied by 5m to the 5th. Now, we can drop off the extra constants we have because as M gets large, those constants can be approximated to just the exponential values. So, we just dropped off our constants, and we have this sum, which is given by 8/15. Multiplied by 1 divided by m squared. Now we'll find the series piece of him. B subm we will compare to 1 divided by m2. Now we know that by the P series. This series converges. This is because our P is the exponent 2, which is greater than 1. Now, let's find our limit. For the limit comparison test, we have to see our limit L. Which equals the limit, as m approaches affinity of a subm divided by BM. This limit should be between 0 and infinity. And so, If this is true, We can then tell That either. The sum of a.m. And BM Both converge. Or diverge together. So now by the P series, we do tell that B subM converges. We have to check our L value first before we can confirm that A7M also converges. So, We have 8/15 multiplied by 1 divided by m2 divided by 1 divided by m2. And this is a limit. M approaches infinity. Now one divided by M2d cancels itself out. Meaning we have the limit of 8/15th, which is just 8/15. This is between 0 and infinity, so we can confirm that a sub-M converges because B sub-M converges. So, if we look at our possible answers, we determined that this answer is answer A. OK, 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 ∞) ((3k³ + 4)(7k² + 1)) / ((2k³ + 1)(4k³ − 1))

Key Concepts

Comparison Test

Limit Comparison Test

Asymptotic Behavior of Rational Functions

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