 Determine the convergence of the series where n equals one to infinity, that was missing there, the natural log of n over n. Now I wanna actually determine this convergence and I'm gonna use the integral test to help me here because what I know is that the integral test, its convergence, the integral test tells us that the convergence of the series will be comparable to the convergence of the integral one to infinity of the natural log of x over x dx, like so. And so this could actually be very helpful for us because we could try to integrate this thing on the right. But in order to use the integral test, we have to make sure our function, which here it's the natural log of x over x, we have to make sure that it is positive decreasing and continuous. Continuity is not such a big deal, right? Is it continuous? And look at the functions involved here. It involves a natural log of x, which is a continuous function. It involves just x itself, which is continuous. And we take the quotient. So as long as we are avoiding division by zero, we'll be continuous. Now we're trying to do the interval one to infinity because that's where this series starts. And so from one to infinity, we never divide by zero. So the answer is yes, this is a continuous function. Is it a positive function? Is it positive? Well, let's consider that for a moment. The number x is always positive. The natural log, well, I should say that x is positive when we're greater than one, right? What about the natural log? Well, the natural log also is positive when we're greater than one. It's negative when between zero and one. So on the interval in question, the natural log is positive, x is positive, the ratio will be positive. So again, the answer here is yes. What about decreasing? It turns out this is where things get a little bit more tricky, right? Because if we just look at the first couple terms of this thing, we could try to plug in like x equals one, x equals two, x equals three, try to get an idea. Middly, we have to start computing natural logs of two and three and such. That might not exactly be obvious. And it turns out that that only gives you the integer values for determining whether it's decreasing or not. We really should be looking at the derivative, right? Because if we take the derivative, what's that real quick? We'll do that calculation. By the quotient rule, we get low d high minus high d low. Square the bottom, here we go. Let's simplify that a little bit. The derivative of the natural log was one over x. That cancels with the x right there. So that's gonna give us a one minus the natural log of x. This sits above x squared. And to determine whether this thing is decreasing or not, we wanna figure out when is, when is the derivative less than zero? Now, the x squared here, since we're greater than zero here, we're actually greater than one. There's no way that thing could ever be zero and x squared's always gonna be positive otherwise. So if you times both sides by x squared, you're just gonna end up with the inequality one minus the natural log of x. When is that thing less than zero? Well, adding the natural log of both sides, we get one is less than the natural log of x. And then exponentiate. If we take the exponential of both sides e and e here, the natural log and exponential cancel out. The exponential function's increasing so we don't have to flip the direction here it's preserved. And we end up that e is less than x. Or if you prefer, we get that x should be greater than e. So the thing is our function is actually only decreasing when x is greater than the number e. But in particular, one is actually smaller than e. So it turns out our function is not always decreasing like the interval test requires. But one thing I wanna mention about the integral test and many other test and convergence we talk about is that, well, yeah, it has some assumptions, right? On the interval one to infinity, we wanna be positive, decreasing and continuous but really these things just have to be eventualities. It just has to be eventually, eventually continuous, eventually positive, eventually decreasing. That is to say the interval one to infinity is somewhat insignificant. We need to pick a starting place, yes, but that starting place can be much later on. Like for example, if you take the number e, the number e right here is approximately 2.7, blah, blah, blah, blah, that's actually enough. Turns out that if we take x to be greater than three, well, three is bigger than e and that will actually take care of it. That when x is greater than three, then, so that is we're on the interval three to infinity. When x is greater than three, the function will have a negative derivative which means the function's decreasing. It was positive when you're bigger than one, so that's still true for three and then it'll be continuous when you're bigger than one which is still true when you're bigger than three. So what we actually wanna do is substitute out here and we wanna use the interval for three to infinity. So that is we're gonna compare the series n equals three to infinity of the natural log of n over n and we wanna compare that, that's comparable to the integral from three to infinity of the natural log of x over x. So the integral test applies here. The integral test is gonna apply only to this improper norm, we go from three to infinity. Now that's not a big deal for us because if it's eventually decreasing, that means that these guys will be convergent but then we can also go back to one here. The starting value of three is somewhat insignificant. It has to n equal, because if the series is convergent when it starts at three it'll be convergent when it starts at one as well. And so even though the integral test doesn't apply when it starts at one, since it applies when it starts to three, then we can determine the convergence or divergence of this series for which then we can then replace that with n equals one and good to go. And that's why I emphasize the eventuality that if our function is eventually positive decreasing and continuous then we can apply the integral test. It doesn't actually have to be continuous decreasing or positive from one to infinity just eventually, right? So our function could be doing something like no, I'm positive, now I'm negative, now I'm positive, now I'm really negative, now I'm positive again, you know, something like this. And then it's like, okay, now I'm gonna get my life together and get something like this. So you might see like this sporadic behavior at the beginning of the function, but it eventually corrects itself into this positive decreasing continuous function that's sufficient to use the integral test. And you're gonna see us being kind of loose about those assumptions in the future because of this, because of this observation. Now let's consider the anti-derivative here of the natural log of x over x. My actually, I would wanna use here u substitution take u to be the natural log of x and hence du equals dx over x for which we can see that's exactly present in the situation. And so the integral then becomes just a u, a u du, and as we also change the bounds, right, you're gonna get the natural log of three on the bottom, you're gonna get infinity still on the top. Like so, find the anti-derivative of this, you're gonna get u squared over two as you go from the natural log of three to infinity and plug in a finite number doesn't have much of a problem. The issue is when you plug this infinite, you're gonna end up with a infinity squared and then everything else will just become infinite as well. So this improper integral will go off towards infinity, which means that it is divergent. So the improper integral is divergent. So then by the integral test, right, good friend, the integral test right here, this implies that the series is likewise divergent. And since this series is divergent when you start at three, we could swap that to n equals one and it'll be divergent when it goes from one to infinity as well. So the final answer is when it comes to this series in question, the original series n equals one to infinity of natural log of n over n, it'll be divergent by the integral test.