42–76. Convergence or divergence Use a convergence test of you...

Question

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.

∑ (from k = 1 to ∞)k⁴ / √(9k¹² + 2)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to determine whether the series, the sum of N equal to 1 to infinity of n cubed divided by the square root of the quantity of 16 multiplied by into the 8th, plus 5 in quantity converges or diverges, and we're given two possible choices as our answers. For choice A, we have convergence, and for choice B, we have diverges. So in order to determine whether our series here converges or diverges, we're going to use the limit comparison test. So we first need to find a series to which to compare our given series. So for that, we're going to look at how our series behaves at large and. So at large values event, if you look at our denominator, which is the square root. Of the quantity of 16 multiplied by the 8th, plus 5 in quantity. For large N, this goes as the square root of the quantity of 16 multiplied by n to the 8th in quantity, which is just equal to 4 multiplied by the 4th. That means that in cubed divided by the square root of the quantity of 16 multiplied by into the 8th. Plus 5 in 20. Goes as In cubed divided by the quantity of 4 multiplied by n to the 4th in quantity, which is just equal to 1 divided by the quantity of 4 nn quantity. So that means that the series that we want to compare to is going to be the sum of 1 divided by N. So, for the limit comparison test, we need to calculate out, which is going to be equal to the limit. As N approaches infinity of A N divided by B N. Where if we have the series, the sum of a of N, where A N in our case is going to be equal to our given series, which is going to be in cubed divided by the square root. Of the quantity of 16 multiplied by end to the 8th. Plus 5 in 20, and for our other series, we're gonna have the sum of Beasts of N. Where we already said that we want to compare this to the sum of 1 divided by N, that means that B N is going to be equal to 1 divided by N. Now we call that the sum of 1 divided by N is a harmonic series, and it diverges. So, for our limit comparison tests, recall that if L is greater than 0 but less than infinity, then Both the sum of Aabin and the sum of Bin must both either converge or diverge. So using these definitions of AN and BN, we see that L is going to be equal to the limit as N approaches infinity. Of the quantity of n cubed, divided by the square root of the quantity of 16 multiplied by N to the 8th, plus 5 in quantity, and that whole quantity is going to be divided by the quantity of 1 divided by N. The next we just want to simplify this expression. That means that this is going to be equal to the limit. As n approaches infinity. Of into the 4th divided by. The square root of the quantity of 16 multiplied by N to the 8 plus 5 in quantity. Next, what we want to do is divide both the numerator and denominator of our expression here by N to the 4th. In doing so, this is going to be equal to the limit as N approaches infinity of 1 divided by the square root. Of the quantity of 16 plus 5 divided by n in quantity. Now as n approaches infinity, 5 divided by the eight approaches 0. So taking this limit, this is going to be equal to 1 divided by the square root of the quantity of 16 plus 0 in quantity, which when we evaluate this expression is going to be equal to 1 divided by 4. So here our value of L is greater than 0 but less than infinity. So whatever one of our series does does, whether it converges or diverges, the other one must do the same. And since we saw that the sum of BN that series diverges, that means the sum of a subn, which is our series that in question also diverges. So that is going to be the answer 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 B. So I hope that this explanation has been useful, and I'll see you in the next video.

42–76. Convergence or divergence Use a convergence test of your choice to determine whether the following series converge.
∑ (from k = 1 to ∞)k⁴ / √(9k¹² + 2)

Key Concepts

Convergence and Divergence of Infinite Series

Comparison Test for Series Convergence

Asymptotic Behavior and Simplification of Terms

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