 In this video, we'll provide the solution to question 4 for the final exam of Math 1220, and we're asked to determine which of the following series is divergent by the divergence test. So remember where the divergence test tells us. The divergence test tells us that if the sequence in play does not converge to zero, then the series is divergent. We're not saying the sequence is divergent because it could actually converge towards like 7 or something. So we have to find a divergence series by the divergence test. Let's just go through these things one by one by one. If we look at the first one, A, we have an N plus one over the natural light of N. That kind of looks hard. I'm going to go to the next one. You know, and that's the thing is we just have to find the right one that's divergent by the divergence test. Now be careful here. There might be more than one series that's divergent, but we have to find the ones that are divergent by the divergence test. If you look at the second one, B, this is a geometric series and its ratio is two-thirds it's convergent. So it's not that one. On the next one, let's see, you have two N over three N cubed plus one. Notice that this thing is bottom heavy. As N goes to infinity, this sequence will go to zero. So whether it's convergent or not, it doesn't matter. The divergence test wouldn't apply here. In this situation right here, same thing, sign of N over N. And if you want to skip it, I mean, we did skip the first one, right? If you want to skip it, we can come back to it. You look at E, this is actually a P series where P equals two, which is convergent. So that's enough to stop us right there. But admittedly one over N squared does go to zero. Same thing with two-thirds to the N, that does go to zero as well. The sequence does. But we knew that we were convergent. And then this last one, this is one you have to be very careful about. This is the harmonic series. Aha! It's divergent. Is it divergent by the divergence test? How do we know the harmonic series is divergent? That actually comes from the P test, which is a derivative of the integral test. If you take the sequence one over N, that does go to zero. So the divergence is not given by the divergence test. So we actually would knock off the harmonic series. That's not the correct answer. So now we have A, O, and D right here. So let's try to get a little bit better hope on what's going on here. So as N goes to infinity for these sequences, sine of N, well there's no such thing as sine of infinity, but sine is bounded between one and negative one. And so when you look at the denominators, N gets bigger, bigger, bigger, bigger, sine just fluctuates up and down, up and down, up and down. It does turn out that our sequence here, this sequence sine of N over N, it does go to zero. We could do a squeeze theorem type argument there because our sequence is bounded above by one over N and it's bounded below by negative one over N and both of these sequences go to zero, therefore the middle sequence goes to zero as well. So if that thing is divergence, the divergence test does not apply. So at this moment, we now see that just by process elimination, the correct answer has got to be A. But let's actually investigate this one to see why it does the sequence not go to zero. Well, so if you look at this one, what's happening on the top, N plus one on the bottom, the natural log of N, as N goes to infinity, you're going to get infinity over infinity. That's a L'Hopital type thing. So we could replace this with one over one over N, if you take the derivative. That simplifies us to be N. And so as N goes to infinity, that actually goes to infinity, right? And so since the sequence doesn't go towards zero, the divergence test shows us that it must be divergent. So while for a fact, we knew, right, the harmonic series is divergent. The divergence test does not apply there. And so it turns out that choice A is the only series which is divergent by the divergence test.