 let us determine the convergence of the series, n equals one, the sum of n equals one to infinity of n times e to the n, the negative n squared there. Now, we could try a couple of things here. We could maybe try the ratio test. The ratio test might be okay here because we have some products and some exponentials here. We could also try the root test, but if you're going down that direction, the ratio test would probably be a little bit cleaner. And one could try that. If we were to do the ratio test, we're using the sequence, well, we have to look at the sequence a n plus one over a n, consider its absolute value. This we get, well, let's start with a n on the bottom. We're gonna get n times e to the negative n squared. We're just looking at the sequence that we're adding up here when it comes to the ratio test. In the top, we get n plus one times e to the negative n plus one squared inside absolute value. Let's break this thing up. So we're gonna get an n plus one term over an n term. And then times this with the exponentials. We're gonna get e to the n squared on top. Since we have these negative exponents, I'm kinda swapping the location of the exponentials there. The x when it was negative. And then the domino will get e to the n squared plus two n plus one. So notice there, I foiled out the n plus one there. And so this gives us that e to the n squared will cancel with the n squared on the bottom. And so our sequence looks more like the following now. We're gonna get n plus one over n. I dropped the absolute values because as n has to be greater than equal to one, n plus one over n will always be a positive quantity. And so then we also get one over e to the two n plus one here, because the exponential on the top disappears. Also, I dropped the absolute values because everything in question here is positive. So now let's take the limit here. We're taking the limit as n goes to infinity. Now the first one, it's a balanced rational expression n over n, that thing is gonna converge towards a one. So we get a one right there. And for the denominator, as the one, and the one in the numerator is gonna stay a one, right? I said the denominator, I'm sorry. It's looking at the second one, focusing here on the denominator. That's the interesting part. You're gonna get this e to the infinity going on there. So the bottom would be infinity. So this looks something like one over infinity, which is really just a shorthand for saying zero. This is the limit of the ratio sequence. Now by the ratio test, we wanna compare this limit to the number one. If it's less than one, like we see right here, then that actually tells us that the series is convergent. But also we get the benefit of saying it's actually absolutely convergent. Absolutely convergent, which is kind of nice. We like absolute convergence. It's a little bit stronger than convergence. And so we can actually conclude this by the ratio test. And so that's a perfectly good way of determining the convergence here. We get absolute convergence using the ratio test. I like that as a possibility. But it turns out one could also, I think another useful strategy here. Let me rewrite the series n equals one to infinity. n times e to the negative n squared here. Another good candidate would be the integral test here. Because if you take the function f of x to equal x e to the negative x squared, like so, there's actually a fairly nice antiderivative to be working with. We wanna integrate from one to infinity here, x e to the negative x squared dx. We can actually pull off some type of u substitution, take u to equal negative x squared. So du equals negative two x dx for which we need a negative two right here. So we'll divide by negative two to compensate for that. Then we switch our integral to be negative one half the integral, we're gonna end up with e to the u du. Changing our bounds here, when x equals one, u will become negative one. And when x goes off towards infinity, u will become negative infinity, like so. The antiderivative e to the u, actually, let me actually switch the bounds right here cause we do have this negative sign in front anyways. So switching the bounds, we end up with one half the integral from negative infinity to negative one of e to the u du. And therefore, if we go with this, the antiderivative e to u is itself. So we get e to the u over two as we go from negative infinity to negative one. If we plug in negative one, we're gonna end up with one over two e. And then with the other one, we're gonna end up with zero as x goes or as u goes to negative infinity, e to the u would go to zero in that case. So dropping the zero, we get a finite value here. So this also gives us like we saw with the ratio test because that integral turned out to be a finite value that integral was convergent that also tells us that we're convergent here. Now one might think that, well, the ratio test gives absolute convergence. What do we have here? Well, since our series consists only of positive terms, when n is greater than equal to one, n will be positive and e to any power is always positive. So this is already a positive series. So if it's convergent, it'll happen infinitely. If it's absolutely convergent. So the extra word absolutely is not really revealing here, but it is true that it's absolutely convergent. Now this gives you two ways of approaching the same one. I think either one is pretty good. If it was personally me, I liked the ratio test a little bit better here because a ratio test only requires I compute a limit. The integral test requires I compute an anti-derivative, which given the two things, I think limit calculations are generally easier than anti-derivative calculations, but either one would be perfectly good here. Personally, I like to reserve the integral test for one or two cases. It's either my fail safe when I've tried everything else and nothing seemed to be profitable or if I see an obvious anti-derivative. And this one right here, the n times e to the negative n squared, it really does feel like, oh yeah, I do a nice U substitution there. So I might actually be inclined to do the integral test pretty quickly on here, but I think the ratio test does give you a cleaner, shorter argument.