Problem 12 Medium Difficulty

Determine whether the series converges or diverges.
$ \displaystyle \sum_{k = 1}^{\infty} \frac {(2k - 1)(k^2 - 1)}{(k + 1)(k^2 + 4)^2} $

Answer

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Video Transcript

Let's use the Lim comparison test. Let's call this term here a k and then let's call B k T B two. Okay, Cube And then here, Kate in the fifth. So I just get these powers by looking at the largest power in the numerator a three. And then here I have four, and then one more. That's a five. And then I could simplify this if I wanted to. And we know that there's some converges. This is just a P Siri's with P equals two. So now let's use limit comparison. So we look at the limit. Kay goes to infinity, a k a. Over b. K. So that's here's r a K and then we'LL go ahead and multiply this bye bye flipping B and then here let's go ahead and multiply this out. So, of course, here we could also cancel three of those cases as we did before. I should have used this. So in the numerator, Scott and more supply this all outlets foil this. That's two K and then we have to the fifth and then here, minus king of the fourth and then minus two cake cute plus case. Where in the denominator king of the fourth to face Where? Plus one And then let's just go ahead to the side and simplify this denominator that'LL be killing us is two k cube plus que plus Kato The fourth two case Where plus one So that's in our denominator. And then here we're almost done after we've right this denominator, Let's go ahead and divide top and bottom. Bye, kid of the fifth. So here, let's see Divide by Kato the fifth. So this is once on the numerator and also on the denominator. So divide this by King of the Fifth and I'LL write this all the way over here We'LL have limit two minus one over Kay minus two over. Case Weird plus one over K Cube. That's our numerator and the denominator one plus one over. Okay, plus two over Kate Square plus two over K cute plus one over Kato, the fourth plus one over Kate in the fifth. Now, as we take the limit, all of these fractions with the K and the denominator go to zero and we're just left over with two divided by one. So at this point, we can go ahead and used what your book calls the limit comparison test. So we've computed the limit and we got a number that's between zero and infinity, so we can go ahead and use this test by the limit comparison. This Siri's one in our question. That should be okay, not a end. And here, and that's our final answer.