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 ∞) 20 / (∛k + √k)

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says, does the series, the sum of M equals 1 to infinity of 8 divided by the quantity of m raise to the quantity of 1 divided by 5 in quantity, plus M raised to the quantity of 3 divided by 4 in quantity converge or diverge. Use the direct comparison test or the limit comparison test. 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 the series that we're given, we notice that each term in the series is going to be a positive value. So that means that the direct comparison test is going to be the most convenient test to use. So recall for the direct comparison test that for positive sequences AM and BM, that if L, which is equal to the limit. As M approaches infinity of a m divided by BM exists, and 0 is less than L, which is less than infinity, then the series the sum of Ace of m and the series the sum of BM, they both either need to converge or diverge. So the next step is to define our sequences A M and B sub M. Now for Ace of M, this is just going to be the original sequence that we had for our series. So AM is going to be equal to 8 divided by the quantity of M raises the quantity of 1 divided by 5 in quantity, plus M raises the quantity of 3 divided by 4 in quantity. Now if we look at this fraction, And we look at the denominator, we notice that the denominator is going to be dominated by the term of m raised to the quantity of 3 divided by 4 in quantity for large values of M. So that means for B subm we're going to want to choose this to be equal to 1 divided by m raised to the quantity of 3 divided by 4 in quantity. So substituting in these sequences back into our definition for L, we see that L is going to be equal to the limit as M approaches infinity of the quantity of 8 divided by the quantity of M raised to the quantity of 1 divided by 5 in quantity plus M raised to the quantity of 3 divided by 4. In quantity, and this is going to be divided by the quantity of 1 divided by M raised to the quantity of 3 divided by 4 in quantity. And simplified this expression, we see that this is going to be equal to the limit as m approaches infinity of 8 multiplied by m raised to the quantity of 3 divided by 4 in quantity, and this is going to be divided by the quantity of m raised to the quantity of 1 divided by 5 in quantity plus m raised to the quantity of 3 divided by 4 in quantity. And now we want to do is divide both the numerator and denominator by M rates of the quantity of 3 divided by 4. And so this is going to be equal to the limit as M approaches infinity of 8 divided by the quantity of 1 divided by m raised to the quantity of 11 divided by 20 in 20 plus 1. Now, as M approaches infinity, the quantity of 1 divided by M raised to the quantity of 11 divided by 20 in quantity is going to approach 0. So taking this limit, this is going to be equal to. 8 divided by the quantity of 0 + 1, which is just equal to 8. So our limit here does exist, and the value of L, which is 8, is between 0 and infinity. So that means if we determine that either of our series, the sum of A subm or the sum of B subm, either diverges or converges, that means the other one does as well. So let's look at the series. The sum. Of B M. Which is going to be equal to the sum. Of M equal to 1 to infinity. Of the quantity of 1 divided by M in quantity rates to the quantity of 3 divided by 4. So this here is a P series, where P is equal to 3 divided by 4, but since this is less than 1, that means that this series diverges. And since this series diverges, if we look at the series, the sum of Ace of M, which is equal to the sum of M equal to 1 to infinity. Of 8 divided by the quantity of m. Raised to the quantity of 1 divided by 5 in quantity, plus M raised to the quantity of 3 divided by 4 in quantity, that means that this series also diverges. And this here is our original series that we're asked to determine whether it converges or diverges. So the answer to this problem is that that series diverges. And if we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

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

∑ (k = 1 to ∞) 20 / (∛k + √k)

Key Concepts

Comparison Test

Limit Comparison Test

Behavior of Series Terms Involving Roots

Watch next