 In this video we provide the solution to question number six for practice exam number four for math 1220 in which case we're given six infinite series and we have to determine which one is divergent by the divergence test. Now, some of them could be divergent for other reasons, but we have to use the divergence test here which said virgins test shows that a series is divergent when the associated sequence of terms doesn't converge to zero in that situation. So let's look at some of these things. You look at the first one in plus one over the natural log of n. That kind of seems kind of hard. I'm going to go somewhere else. This is a geometric series right here where the ratio is small two thirds actually is going to be convergent. So that's not the answer. This one over here, if you look at this one, the top is a two in the bottom is a three in cubed. So this thing the sequence is approximately two over three in squared as n goes to infinity that's going to go to zero. So that could be convergent, it could be divergent. I don't know, but it's not divergent by the divergent test. Over here, sine of n over n. Well, sine is going to keep on going back and forth between one and negative one as n gets bigger. That thing's going to go down towards zero. So that sequence converges to zero is the series convergent. Don't know, don't care. This one right here, this is actually a convergent P series throw that one out. And then the last one is the harmonic series for which we know the harmonic series is divergent. But if you look at the sequence, the sequence of terms one over n does converge towards zero. And so while this is divergent is not divergence by the divergence test. And so we wouldn't include that one. So honestly, by process of elimination, I know that the correct answer has to be a. Now, if we actually want to see the calculation limit as opposed to this elimination method that we did here, let's actually go through it the limit as n goes to infinity of n plus one over the natural log of n. This currently has an indeterminate form infinity over infinity. So we'll apply L'Hopital's rule here to see what's going on. In the top, you're going to get back a one. In the denominator, you're going to get one over n as n goes to infinity. So I would actually want to times the bottom by n, I got times the top by n as well. And so this turns in the whole ratio into the limit as n goes to infinity. Your ratio is actually going to become n now. And as n goes to infinity, this goes off towards infinity, which notice is not zero. So that itself, we actually computed the limit from the beginning. We actually could have seen that this limit, the limit of the sequence doesn't go towards zero. So it is divergent by the divergence test. But process of elimination is an appropriate tool to use on this one as well.