Limit Comparison Test
In Exercises 9–16...

Question

Limit Comparison Test

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

∑ (from n=1 to ∞) n(n + 1) / ((n² + 1)(n − 1))

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 whether the series, the sum evaluated from n is equal to 1 to infinity of. N multiplied by parentheses of 3 n + 1 divided by parentheses of n to the power of 2 + 4 multiplied by parentheses of n + 2 converges or diverges using the limit comparison test. OK. So it appears for this particular prompt, we're asked to determine whether or not the series that is provided to us will converge or diverge 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 and apply what exactly is the limit comparison test. So based on our prior knowledge, we should be able to recall that the limit comparison test states that if we let A N be greater than 0 and bn be greater than 0, for all n, specifically 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 0, infinity parenthesis. And if 0 is less than l and L is less than infinity, then the sum of a n 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 b n diverges, then the sum of a n diverges. Otherwise, the test will be inconclusive. So keeping in our rule, keeping in mind our rules for the limit comparison test, we now need to apply. A subscript NAN for short, will be equal to n multiplied by parentheses of 3n plus 1. Divided by parentheses of n2 + 4 multiplied by parentheses of n + 2. And we need to note that for n is greater than or equal to 1, we will have A N is greater than 0. And we need to compare with BN is equal to 1 divided by n. Because for a large value of N, the given term behaves like a constant multiplied by 1 divided by N. We also need to note, so let's make a note that the sum evaluated from n is equal to 1 to infinity of 1 divided by n. Is a divergent, which I'll put D in quotes, is a divergent P series, since P is equal to 1 and 1 is less than or equal to 1. So now we need to compute the ratio, so we need to note that AN divided by BN. Will be equal to n multiplied by parentheses of 3 n + 1 divided by parentheses of n2 + 4 multiplied by parentheses of n + 2 divided by 1 divided by n which will be equal to n to the power of 2 multiplied by parentheses of 3 n plus 1. Divided by parentheses of n to the power of 2 + 4 multiplied by parentheses of n + 2. Fantastic. We are on a roll. So, we now need to find the limit, so L is going to be equal to the limit as N approaches infinity of. So it's going to be as the limit as n approaches infinity of n to the power of 2 multiplied by parentheses of 3 n + 1 divided by parentheses of n to the 2 + 4 multiplied by parentheses of n + 2. Which will be equal to the limit as n approaches infinity of 3 multiplied by n to the power of 3 plus n to the power of 2 divided by n to the power of 3 + 2 multiplied by n to the power of 2. + 4 and + 8. And since the numerator and denominator have the same degree, that will mean that the limit is equal to the quotient of the leading coefficient, meaning that l will be equal to 3 divided by 1, which is equal to 3. And since 0 is less than 3 and 3 is less than infinity. And since the sum of, so the sum evaluated from n is equal to 1 to infinity of 1 divided by n. Diverges that will mean that the series that is provided to us by the prom itself will also diverge. So that means our final answer is diverges hooray, we did it. So going back to the top to recap what we've learned and to look at the prom itself. We could safely say, based on what we've learned together, that our final answer without a doubt will be diverges. So our series without a doubt will diverge. 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 Test
In Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.
∑ (from n=1 to ∞) n(n + 1) / ((n² + 1)(n − 1))

Watch next

Limit Comparison TestIn Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.∑ (from n=1 to ∞) n(n + 1) / ((n² + 1)(n − 1))

Watch next

Limit Comparison Test
In Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.
∑ (from n=1 to ∞) n(n + 1) / ((n² + 1)(n − 1))