 Let us determine the convergence of the given series, the sum, where n ranges from one to infinity, of the ratio with the square root of n cubed plus one as its numerator and as its denominator, three n cubed plus four n squared plus two. So take a moment, maybe a minute, to think about on your own, which test of convergence or divergence would you use to determine the convergence or divergence of the series? Pause the video right now if necessary. Now, if it was me, I noticed we have a rational expression. We have a bunch of polynomial expressions like an n cubed on top, an n cubed on bottom. We also have this square root going on here. This thing is kind of messy in terms of algebraic operations. I am very inclined to run a limit comparison test. With all the pluses on top and bottom, it'll be very difficult if at all possible to get a legitimate comparison test to work. So I would be inclined to try limit comparison test. And I think that because again, it feels so messy. Like that's the official word in consideration here. It looks kind of messy. There's all this stuff going on because the other convergence test that I might be tempted to do will be like the integral test, but I don't want to find the anti-derivative of that thing. I think comparison test is going to work out great. If we just look at the leading terms on top and bottom, our series is going to look a lot like the following. I believe that it's approximately the same thing as the series where n equals one to infinity. That part will be the same. We're going to get a square root of n cubed on top. Don't forget the square root and a three n cubed on the bottom. In terms of exponents, we have n to the three halves on top, we have n to the three on the bottom. And so by exponent rules, we would subtract the exponents and end up with the series n equals one to infinity of one over three times n to the three halves power. And so this, we're comparing our series to a p-series. So the original series, which we see at the top there, it's comparable to this p-series. So we're going to use the p-test on this one where our p-value is equal to three halves. A p-series whose p-value is greater than one will be convergent. So this p-series right here, the multiple of three will have no effect on it whatsoever. This p-series is convergent. It's convergent by the p-test. That tells us this series is convergent. But then the limit comparison test is going to tell us that the convergence of this series right here will be the same convergence as this series right here. So this series will likewise be convergent and this follows by the limit comparison test. I really like to use the limit comparison test when my sequence in the series is super messy. I think this one constitutes such an example and that would be my recommendation.