Limit Comparison Test
In Exercises 9–16, use the Limit C...

Question

Limit Comparison Test

In Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.

∑ (from n=2 to ∞) 1 / ln n

(Hint: Limit Comparison with ∑ (from n=2 to ∞) (1/n))

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine convergence or divergence of the series, the sum evaluated from n equals 2 to infinity of 1 divided by the natural log of n to the power of 2 using the limit comparison test. OK. So it appears for this particular problem, we are given a series, and we're asked to determine whether or not this series will converge, either A converges or B diverges by using the limit comparison test. So now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to recall what we know about the limit comparison test. So based on our prior knowledge, which you should be able to access this information easily in your textbook, or you can quickly look it up online, but the limit comparison test states that if we let n be greater than 0 and bn be greater than 0, for all lowercase n is greater than or equal to N, where N is an integer. And assume that the limit l is equal to the limit as n approaches infinity of a n divided by bn exists in square bracket of 0, infinity. If 0 is less than L and L is less than infinity. Then the sum of an and the sum of bn either both converge or both diverge. And if l is equal to 0 and the sum of bn converges, then the sum of a n converges, and if l is equal to infinity and the sum of bn diverges, then the sum of a n diverges. Otherwise, the test will be inconclusive. So keeping the limit comparison test rules in mind, we now need to apply to a subscript NAN for short. Which will be equal to 1 divided by parentheses of the natural log of n to the power of 2. Fantastic, and we need to note for n is greater than or equal to 1, we will have that AN is greater than 0, and we need to compare with BN and BN is equal to 1 divided by n. And this is where the sum evaluated from n equals 2 to infinity of 1 divided by n is divergent, which I'll put a d in quotes. Once again, this will be a divergent P series since P will be equal to 1 and 1 is less than or equal to 1. So now we need to compute the ratio, so when you take A N divided by BN. Which will be equal to 1 divided by parentheses of the natural log of n to the power of 2 divided by 1 divided by n, which will be equal to n divided by parentheses of the natural log of n to the power of 2. So now we need to find the limit, so L will be equal to. The limit as n approaches infinity of n divided by parentheses of the natural log of n. To the power of 2, which will be equal to square brackets of infinity divided by the parentheses of the natural log of infinity to the power 2, which will be equal to square brackets of infinity divided by infinity. And as we should recall and be able to recognize, infinity divided by infinity is an indeterminate answer. And that means we must recall and apply Lahapatal's rule, and in order to apply Lahapatal's rule, we will need to differentiate the numerator and denominator separately with respect to n. And after applying the Lahapatal's rule, if the expression still results in an indeterminate form, that will mean we will need to repeat the differentiation process until a determinate form is achieved as an answer. So that means we need to go ahead and write that L is equal to the limit as n approaches infinity of d by DN of N. Divided by d by dn of square brackets of parentheses of the natural log of n to the power of 2. Which will be equal to the limit as n approaches infinity of 1 divided by 2 multiplied by parentheses of the natural log of n multiplied by 1 divided by n. And this will be equal to the limit as n approaches infinity of n divided by 2 multiplied by the natural log of infinity, which will be equal to square brackets of infinity divided by, so it's gonna be infinity divided by 2 multiplied by the natural log of infinity. Which will be equal to. Square brackets of infinity divided by infinity, which unfortunately is an indeterminate answer, so we need to apply Lahapatal's rule again. So we need to take l is equal to the limit as n is equal, or rather l is equal to the limit as n approaches infinity of d by dn of n. Divided by D by DN. Of 2 multiplied by the natural log of n. Which will be equal to the limit as n approaches infinity of 1 divided by 2 divided by n. Which will be equal to the limit. As N approaches infinity of drum roll and divided by 2. And this will be equal to square brackets of infinity divided by 2, which will be equal to infinity. So that means that since l is equal to infinity and the sum b n diverges, that will mean that the sum a n will also diverge. So our final answer is diverges. Hooray, we did it. So looking at the prom itself, we could safely say that our final answer without a doubt will be multiple choice answer letter B diverges, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Limit Comparison TestIn Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.∑ (from n=2 to ∞) 1 / ln n(Hint: Limit Comparison with ∑ (from n=2 to ∞) (1/n))

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Limit Comparison Test
In Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.
∑ (from n=2 to ∞) 1 / ln n
(Hint: Limit Comparison with ∑ (from n=2 to ∞) (1/n))