40–62. Choose your test Use the test of your choice to determi...

Question

40–62. Choose your test Use the test of your choice to determine whether the following series converge.

∑ (k = 1 to ∞) k⁸ / (k¹¹ + 3)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says that the series, the sum of M equal to 1 to infinity of M4th divided by the quantity of M 7 plus 10 in quantity converge or diverge. And we're given two possible choices as our answers. For choice A, we have converges, and for choice B, we have diverges. Now, if we look at our series, we see it's of the form of the sum of a sub M, where a sub M is going to be equal to M 4th, divided by the quantity of M 7th, plus 10 in quantity. And if we look at A of M, we see that it is going to be greater than 0, or values of M that are greater equal to 1. That means that we can use the limit comparison test to see whether the series converges or diverges. So we need to choose a series? To which to compare our given series here. And if we look at large values of M, we'd noticed that A of M goes as M4th divided by M7th, which is equal to 1 divided by M cubed. So we're going to look at the series where BM is equal to 1 divided by Mcubed. And so, the series, the sum of BM is going to be equal to the sum. Of the quantity of 1 divided by M in quantity cubed, which here here is a P series with P equal to 3, and since that is greater than 1, that means that this series converges. So we're comparing it to a series that we know converges. So, for the limit comparance test, we need to calculate L, which is equal to the limit, as M approaches infinity. Of a sub M divided by BM. And if L here is greater than 0, but less than infinity, then whatever one of the series does, either converges or diverges, that means the other series also. So, for our series here that we're comparing, L is going to be equal to the limit, as M approaches infinity. Of the quantity of M 4th divided by the quantity of M 7th, plus 10 in quantity, that whole quantity is going to be divided by the quantity of 1 divided by M cubed in quantity. And simplifying this expression, we see that this is going to be equal to the limit, as M approaches infinity of M 7th, divided by the quantity of M 7 plus 10 in quantity, dividing both the numerator and denominator by M 7th, we see that this is going to be equal to the limit as M approaches infinity of 1, divided by the quantity of 1 plus 10 divided by M7. Now, as M approaches infinity, 10 divided by M to the 7th approaches 0. So when taking this limit, this is just going to be equal to 1 divided by the quantity of 1 plus 0, which is equal to 1. Now, since L here is greater than 0 and less than infinity, that means that both our series, the sum of Ace of m and the sum of BM, both must either converge or diverge. And since we've shown that the sum of B of M already converges, that means that the sum of a of M also converges. So that means that the answer to this problem is that our series, the sum of Ace of M, converges. So that is the answer that we are looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

40–62. Choose your test Use the test of your choice to determine whether the following series converge.
∑ (k = 1 to ∞) k⁸ / (k¹¹ + 3)

Key Concepts

Convergence of Infinite Series

Comparison Test and Limit Comparison Test

Asymptotic Behavior of Terms

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