 Hello and welcome to a screencast today about using the limit comparison test to see if a series converges or diverges. Alright, here I put up a statement of the limit comparison test just in case you've forgotten and it's always good to remind you of that. So let a sub k and b sub k be series with positive terms. And you notice I put the word positive here in bold. If we know that the limit of as k goes to infinity of b sub k over a sub k gives us some constant c. c has to be a positive and finite number. So finite just means something that's not infinity. Then a sub k and b sub k, the series, either both diverge or they both converge. Okay, so it just depends on what the one's doing means the other one will behave the same way. Alright, so let's take a look at our example here then. So this one says use the limit comparison test to determine the convergence or divergence of the series. And here we've got the series of 6 plus 5 to the n over 3 to the n as n goes from 1 to infinity. Okay, so the first thing probably to figure out, so let's just call this our a sub k or in this case a sub n since I've got n's in here instead of k's. We've got to figure out what we're going to compare it to. Well, let's see, so as n gets really big is this 6 in the numerator here really going to make that much of a difference? Not especially. Okay, so let's call b sub n then the series as n goes from 1 to infinity of 5 to the n over 3 to the n. Okay. Alright, so now that we know what we're going to be comparing it to, and again hopefully you can kind of see where I picked this series from, now we've got to check our conditions. So our first condition is are the series positive? And yes, I would definitely say they're positive. I see only positive numbers, I see only addition, and I see powers of positive numbers. So yeah, these definitely are both positive. Okay, check. Alright, second thing we need to check then is if we do the limit of 1 divided by the other one, and honestly it doesn't matter which way you want to divide them. Probably doing a sub n over b sub n is going to make more sense, although which way did I define it? I said b sub k over a sub k. Okay, so then we could just switch them around, whatever. It doesn't really matter. It just means the limit of when you divide them out. Okay, so let's take a look at that. So let's look at the limit as n goes to infinity. So I'm going to go ahead and stick my a sub n in the numerator. So 6 plus 5 to the n over 3 to the n, and then I'm going to divide that whole thing by 5 to the n over 3 to the n. Okay, so this is just one big mess. But you notice we have a fraction over a fraction. So what I'm going to do first then is I'm going to go ahead and multiply by that reciprocal of my denominator. So I've got 6 plus 5 to the n over 3 to the n. I'm going to flip this denominator then. So I'm going to end up with 3 to the n over 5 to the n. Okay, you'll notice something magical happens, right? Our 3 to the n's cancel. That's wonderful. So now I'm going to end up with the limit as n approaches infinity of 6 plus 5 to the n over 5 to the n. Okay, well, again, is the 6 really going to make a lot of difference? No, but I'm not totally convinced yet. So let's break up that fraction. So let's do the limit as n goes to infinity, and let's break it up because now we only are multiplying, or we're dividing by one piece in the denominator. So I'm going to do 6 to the 5 over n plus 5 to the n over 5 to the n. Okay, well, the second piece here is going to give us 1, so that's a nice number. And now I'm convinced. I definitely agree that as n goes to infinity, this denominator is going to blow up, right? So this whole first piece here of my limit is going to end up going to 0. Okay, so that's going to give us 1. 1 is definitely a positive finite number. Okay, so we've got our second condition confirmed then. Okay, so now we either have to decide whether they're both going to converge or they're both going to diverge. Well, you'll notice that. Okay, so now we've got to ask converge or diverge, and I'm abbreviating these a little bit. The series that I picked here from my b to the n, that looks vaguely familiar. Okay, do you guys agree that I could rewrite this as n goes from 1 to infinity of 5 over 3 to the n? Everybody buy that? I'm just kind of using rules of exponents backwards, because typically we distribute an exponent in. Here I'm kind of factoring it out, right? And that's legit because I've got a fraction here that I'm working with. Okay, well, this is a geometric series. This is why we've been harping on these things for so long because we want to make sure that you guys can recognize these things. In this case, my r is going to be 5 thirds, okay, which is definitely greater than 1, so that means it's going to diverge. Okay, so that means then, I'm going to go ahead and finish my question. So that means our original series, n goes from 1 to infinity of 6 plus 5 over n to the 3n also diverges. And that was the original question that we needed to answer, okay? So we had to check a couple of initial conditions until we got here, and then we had to make sure we had a series that we actually recognized, which we did, that one diverged, so this one also diverges. All right, thank you for watching.