 In this video, we're gonna use the integral test to define the convergence of a series. So consider the series here, n equals one to infinity of one over n squared plus one. Is this thing convergent or not? So now to use the integral test, we have to compare the sequence in play with, which is a discrete function there with the continuous expansion of it. So notice here that we wanna consider the function f of x equals one over x squared plus one, like so. And then the sequence in play, a sub n, this is equal to one over n squared plus one. And just as a distinction here, when we talk about the sequence, we'll often use subscripts to represent the variable and then the number in play here is gonna be an n for the natural number n, as opposed to the continuous function f of x for which we'll use an x to denote the variable here. Now these two things are related to each other. And we wanna use the integral test because this f of x is the expansion of the sequence a sub n, which is the same sequence that we picked up right here. Now in order to use the integral test, we need three things. First, we need to check is the sequence or is the function positive, is f a positive function? That is, if we choose any number, will f of x be greater than zero? Now let's start off with the number x, right? Well, if we square it, square any real number, that's gonna be greater than or equal to zero. If you add one to it, like here x squared plus one, that's always gonna be greater than zero, right? And then when you take the reciprocal of a positive number, that still is gonna be positive. So one over x squared plus one will be greater than zero. And this is true for any choice of x here. So the answer is yes, f of x is always positive. Absolutely. It's gonna be positive for all real numbers, negative infinity to infinity. Now for the case of the integral test, we only need this to be true for one to infinity. So in particular, since it's positive everywhere, it's gonna be positive on the interval one to infinity. That's the first one, great. Let's check for continuous. This is often an easy one to check, but is it continuous, we need it? Well, take our function here, one over x squared plus one. We're gonna see that this thing is continuous because if you take the number x and you square it, square it as a continuous function, we add one to it, adding as a continuous function. And if you divide by one, that is to take one divided by x squared plus one, division is a continuous operation. The only issue we have to worry about is does the denominator ever go to zero? Now, as we observed before, x squared plus one is always greater than zero. So the denominator here can never equal zero. So there's no discontinuities, no vertical asymptotes, no removal discontinuities. We're continuous, right? So because the x squared plus one is positive, that we don't have to worry about discontinuities. That's the only thing we have to worry about there. So then that brings us to the last topic here about is it decreasing? That's the last assumption one needs for the integral test. Now to see that it's decreasing, what I want you to do is compare the following. On the interval one to infinity, take the expression x squared. And well, maybe not x squared, we should be talking about the sequence right here. Take n squared instead, sorry about that. So if n squared, it's less than or equal to n plus one squared. I'm just saying here that the squaring function is an increasing function. Now when you take the reciprocals of this, one over n squared will actually be greater than one over n plus one squared. If numbers are getting bigger, the reciprocals are gonna get smaller. And so this tells us that our function is decreasing. Our function is decreasing. Now if we wanted, so this is actually showing that the sequence is decreasing, which is all that we really need here. But the function, we also get that it's decreasing. You could use the derivative to help you out here if there's any concern whatsoever. But since our function is positive, continuous, and decreasing, what we can do is we can see that our series will be convergent if and only if the integral is convergent. So we wanna look at the improper integral one to infinity of the function one over x squared plus one dx. And now with this integral, we can use the fundamental thing with calculus. For one over x squared plus one, we might think of doing a trig substitution, which would actually be right. We could take x to be tangent theta, but this actually is a very common form. One over x squared plus one is the derivative of arc tangent. So if we did that trig substitution in the end, we'd end up with arc tangent of x there. So I'm kinda skipping the steps there as we go from one to infinity. And so plugging these things in here, we take arc tangent of infinity minus arc tangent of one. We see right here. Now, when I say arc tangent of infinity, I don't actually mean we plugged infinity into arc tangent because infinity's not a number, so we can't actually evaluate it there. This, of course, is just an abbreviation for the limit. The limit has x approaches infinity. What happens to arc tangent of x? Now, because arc tangent has a horizontal asymptote, this asymptote actually is pi halves. And arc tangent of one, that's just pi force. This thing would combine just to give us pi force. This tells us that the integral, this improper integral actually adds up to be pi force. It's less than infinity. This is evidence that the integral is convergent. And we're talking about the improper integral is convergent. Therefore, the integral test comes into play right here that the convergence of the integral actually implies, where was it? There you are. The series likewise is going to be convergent. And why is that? Because when we talk about the convergence and divergence of series, it's not just good enough to say convergent, divergent. You need to provide the evidence on why it's convergent. And the evidence comes from the integral test. Because the improper integral was convergent, the associated series must likewise be convergent. Because if the integral was divergent, then the series would be divergent. If the integral is convergent, then the series has to be convergent as well.