 Let's determine the convergence of the series, where we take the sum n equals 1 to infinity of 1 over 2 plus 3 to the n. Take a moment just by yourself, pause this video if necessary, and think about which convergence test would be most applicable for a series like this. Now if it was me, I would think about something like the following. If I'd look at just the top, we got a 1, not a whole lot going on there, but if it's nominated the leading term is a 3 to the n. Now because of that, this thing, I feel like this sequence 1 over 2 plus 3 to the n, this would be asymptotic to the sequence 1 over 3 to the n, which this is a geometric sequence, 1 third to the n, where your constant ratio is 1 third, and so that would imply that the associated geometric series would be convergent. So I feel like I should use a comparison test where I compare this to a geometric series, for which case you can use the limit comparison test, use the limit comparison test, in which case we'd have to evaluate the limit as n goes to infinity of the sequence 1 over 2 plus 3 to the n over 1 over 3 to the n. You want to simplify that thing, you end up with the limit as n goes to infinity of 3 to the n over 2 plus 3 to the n here, and so kind of like we mentioned before, we have this balanced term here, this 3 to the n 3, and those are the dominant terms on top, and so as we go towards infinity, you're going to end up with just these, you're just going to get the limit of 3 to the n over 3 to the n, that would cancel, given it's just a 1. So the limit comparison test would be applicable here, and so since the series 1 over 3 to the n, I should take the sum from 1 to infinity here. This was a geometric series, so the geometric series test applies. This tells us that this associated series is convergent because our ratio was small, and therefore by the limit comparison test, the limit comparison test, we then get convergence of the original series. That is a perfectly great way of handling this one, right? Now the limit comparison test is very versatile in that regard, but the problem is it requires a limit calculation, which in this case, it's really not that bad. I can kind of, I mean, personally, a lot of these limit calculations are ones we can just do in our head, or we can sort of eyeball it, whatever that means. But, you know, if you're a student in a classroom, you might be expected to provide all the details of that, and I kind of skipped some of the details on that last limit there. So if you need to show a little bit more detail, the limit comparison test might be a little bit less preferable. It turns out in this situation, the genuine comparison test actually works out pretty nicely here, the actual comparison test, because if you remove the plus two in the denominator, you've made the denominator get smaller, which makes the fraction actually get bigger. So this series is actually comparable to n equals one to infinity of one over three to the end, which is then a geometric series, which is convergent because its ratio is small. So you get that this is likewise convergent. And so the comparison test is a little bit more preferable here because it doesn't require a limit calculation. But with enough experience, that limit calculation we did, which we saw on the screen a moment ago, is really not that taxing. So a comparison test or the limit comparison test are what I would recommend for this, for this series.