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# Integral test

AP Calc: LIM‑7 (EU), LIM‑7.A (LO), LIM‑7.A.6 (EK)

## Video transcript

- [Voiceover] Let's explore a bit the infinite series from n equals one to infinity of one over n squared. Which of course is equal to one plus one fourth, that's one over two squared, plus one over three squared, which is one ninth, plus one sixteenth and it goes on and on and on forever. So there's a couple of things that we know about it. The first thing is that all of the terms here are positive. So all of the terms here are positive. So they're all positive and that they're decreasing. It looks like they're decreasing quite quickly here from one to one fourth to one ninth to one sixteenth, and so they're quickly approaching zero, which makes us feel pretty good that this thing has a chance of converging. And because they're all positive we know that this sum right over here, if it does converge is going to be greater than zero. So the only reason why it wouldn't converge is if somehow it goes unbounded towards infinity, which we know if this was one over n it would be unbounded towards infinity. So this says that's a possibility here. So if we could show that this is bounded, then that will be a pretty good argument for why this thing right over here converges because the only reason why you could diverge is if you went to either positive infinity or negative infinity. We already know that this thing isn't going to go to negative infinity because it's all positive terms. Or you could diverge if this thing oscillates, but it's not going to oscillate because all of these terms are just adding to the sum, none of them are taking away because none of these terms are negative. So let's see if we can make a good argument for why this sum right over here is bounded, especially if we can come up with the bound, then that's a pretty good argument that this infinite series should converge. And the way that we're going to do that is we're going to explore a related function. So what I wanna do is I wanna explore f of x is equal to one over x squared. You could really view this right over here one over n squared as f of n if I were to write it this way. So why is this interesting? Well let's graph it. So that's the graph of y is equal to f of x. And notice this is a continuous, positive, decreasing function, especially over the interval that I care about right over here. I guess we could say for positive values of x, it is a continuous, positive, decreasing function. And what's interesting is we can use this as really an underestimate for this area right over here. What do I mean by that? Well one, this first term right over here, you could view that as the area of this block right over here. That is f of n or I guess you can say f of one high and one wide, so it's going to be one times one over one squared or one. Let me make sure I'm using different colors. This term right over here, that could represent the area of this block, which is one fourth high and one wide so it is going to have an area of one fourth. What could this one represent? Well the area of the next block if we're trying to estimate the area under the curve. And this might look familiar from when we first got exposed to the integral or even before we got exposed to the integral and we were taking Riemann sums. So that right over here, that area, is going to be equal to one ninth. So what's intriguing about this is we know how to find the exact area, or the exact area from one to infinity, from x equals one to infinity. So maybe we can use that somehow. We know what this area is right over here, which we can denote as the improper integral from one to infinity of f of x dx. We know what that is and I'll figure it out in a little bit. And if we know what this is, if we can figure out the value that's going to be an upper-bound for one fourth plus one ninth plus one sixteenth on and on and on and on. And so that will allow us to essentially bound what this series evaluates to and as we said earlier that would be a very good argument for its convergence. So the whole point here, I'm not doing a rigorous proof, but really getting you the underlying conceptual understanding for a very popular test for convergence or divergence which is called the integral test. Let me just write that down just so you know what this is kind of the mental foundations for. So what do I mean here? So let me write this sum again. Let me write it a little bit different. So our original series from n equals one to infinity of one over n squared. It's going to be equal to this first block, the area of this first block plus the area of all the rest of the blocks, the one fourth plus one ninth plus one sixteenth, let me do this in a new color. Which we could write as the sum from n equals two to infinity of one over n squared. So I just kind of expressed this as a sum of this plus all of that stuff. Now what's interesting is that this, what I just wrote in this blue notation that's this block plus this block plus the next block, which is going to be less than this definite integral right over here. This definite integral, notice it's an underestimate it's always below the curve, so it's going to be less than that definite integral. So we can write that this thing is going to be less than one plus instead of writing this I'm gonna write the definite integral. One plus the definite integral from one to infinity of one over x squared dx. Now why is that useful? Well we know how to evaluate this and I encourage you to review the section on Khan Academy on improper integrals if this looks unfamiliar, but I'll evaluate this down here. We know that this is the same thing as the limit as, I'm going to introduce a variable here, t approaches infinity of the definite integral from one to t of and I'll just write this as x to the negative two dx. Which is equal to the limit as t approaches infinity of negative x to the negative one, or actually I could write that as negative one over x. And we're going to evaluate that at t and at one, which is equal to the limit as t approaches infinity of negative one over t and then minus negative one over one so that would just be plus one. And as t approaches infinity this term right over here is going to be zero, so this is just going to simplify to one. So this whole thing evaluates to one. So just like that we were able to place an upper-bound on this series. We're able to say that the series under question or in question, so the infinite sum from n equals one to infinity of one over n squared is going to be less than one plus one or it's going to be less than two. Or another way to think about it, it's going to be the two is this area, that's one right over there plus this area right over here. So we're saying that this sum is going to be less than two so we have bounded it above. So we know that it cannot go to positive infinity. Because all the terms are positive it's definitely not going to go to negative infinity. And because all the terms are positive we also know that this isn't going to oscillate between two different values, so this gives us a pretty good sense that this series converges. And the logic we just used here to argue for why this converges, once again not a rigorous proof, but this is the underlying logic of the integral test.