Proof of p-series convergence criteria | Series | AP Calculus BC | Khan Academy

You might recognize what we have here in yellow as the general form of a p series. What we're going to do in this video is think about under which conditions, under what p, will this p series converge. By definition, for it to be a p series, p is going to be greater than zero.

I've set up some visualizations to think about how we are going to understand when this p series converges. So over here you have the graph; this curve right here, that's y is equal to 1/x to the p. We're saying it in general terms because p is greater than zero. We know it's going to be a decreasing function like this; once again, that's y is equal to 1/x to the p.

Now what we've shaded in ahead of time underneath that curve, above the positive x-axis, that is the integral from 1 to infinity, the improper integral of 1/x to the p dx. So that's this area that I have already shaded in; you see it in white in both of these graphs. What we're going to hopefully see visually is that there's a very close convergence or divergence relationship between this p series and this integral right over here.

Because when we look at this left-hand graph, we see that this p series can be viewed as an upper Riemann approximation of that area. What do I mean by that? Well, think about the area of this first, I guess you could say this first rectangle; the width is one and its height is 1 over 1 to the p. So this would be the first term in this p series. This would just be an area of one; just the scales on the x and y axes are not the same. This one right over here, its area would be 1 over 2 to the p. This area is 1 over 3 to the p.

So the sum of the areas of these rectangles—that is what this p series is. You can see that each of these rectangles is covering more than the area under the curve. And so we know the area under the curve; that's going to be greater than zero. This p series is going to be greater than this integral, greater than the area under the curve. But if we add one to the area under the curve, so now we're not just talking about the white area; we're also talking about this red area here, well then our p series is going to be less than that.

Because the first term of our p series is equal to one, and then all of the other terms you can view as a lower Riemann approximation of the curve, and you can see they fit under the curve and they leave some area. So this is going to be less than that expression there. Now think about what happens if we know that this right over here diverges.

So if this improper integral diverges, it doesn't converge to a finite value. Well, this p series is greater than that. So if this diverges, then that's going to diverge. Similarly, if this converges—the same integral right over here—if this converges, it goes to a finite value, well one plus that is still going to converge. And so this p series must also converge; it must go to a finite value.

And all I'm talking about right here is really just the integral test. When we think about tests of convergence and divergence, I'm just making sure that we have a nice conceptual understanding and not just blindly applying the integral test. You could go the other way too; if the p series converges, then for sure this integral is going to converge. If the p series diverges, then for sure this expression right over here is going to diverge and the integral diverges.

So we can say the p series converges if and only if this integral right over here converges. So figuring out under what conditions, for what p, does the p series converge, is boiling down to under what conditions does this integral converge. So let's scroll on down to give us some real estate to think about what has to be true for that integral to converge.

I'm going to rewrite it. So we got the integral from 1 to infinity, the improper integral of 1/x to the p dx. This is the same thing; this is the limit as I'll use the variable M since we're already using n as M approaches infinity, and the integral from 1 to M of 1 over x to the p dx.

Let me just focus on this and we'll just remember that we're going to have to take the limit as M approaches infinity. I don't want to have to keep writing that over and over again. So let's think about what this is. So there's a couple of conditions we know; we already know that p is greater than zero. We know that p is greater than zero, but there are two situations right over here.

There's one situation when p is equal to 1. If p is equal to 1, then this is just the integral of 1/x. So this thing is going to be the integral of ln of x, and we're going to go from 1 to M. So this would be the natural log of M minus the natural log of one.

Well, e to the 0 power is one, so the natural log—I’ll write it out—the natural log of one, but the natural log of one is just zero. So when in the special case, I guess we can say when p equals 1, this integral from 1 to M comes down to the natural log of M. Now let's think about the situation where p does not equal one. Well, there, we're kind of just reversing the power rule that we learned in basic differentiation.

So we'd increment that exponent, so it would be x to the p + 1, and we could even write that as x to the 1 minus p. That's the same thing as p + 1, and then we would divide by that, so 1 minus p. We are going to go from 1 to M. This is going to be equal to; we could write this as m to the 1 - p over 1 - p minus 1 to the 1 - p over 1 - p.

So now let's take the limits. So we remember this integral; we want to take the antiderivative or the definite integral here, but then we want to take the limit as M approaches infinity. So what is the limit as M approaches infinity of the natural log of M? Well, if M goes unbounded to infinity, well, the natural log of that is still going to go to infinity. So when p equals 1, this thing doesn't converge; this thing is just unbounded.

So p equals 1, we diverge; we know that. Now let's look over here; let's think about the limit as M approaches infinity of this expression right over here. The only part that's really affected by the limit is the part that has M. So we could even write this as we could take this 1 over 1 minus p out of this. We could say 1 over 1 minus p times the limit as M approaches infinity of M to the 1 minus p, and then separately we can subtract 1 to the 1 minus p.

Well, for any exponent, that's just going to be one. So one over 1 minus p is that right? Yeah, no matter what exponent I put up here, one to any power is going to be one. And so the interesting thing about whether it converges or not is this part of the expression right over here, and it's all going to depend on whether this exponent is positive or negative.

If 1 minus p is greater than zero, if I'm taking that thing to a positive exponent, well then this is going to diverge. In this situation, we diverge, and 1 minus p is greater than zero; we can add p to both sides. That's the situation where 1 is greater than p, or p is less than one; we are going to diverge.

So so far we know that p is going to be greater than zero, and if p is one or if it's less than one, we're going to diverge. But if this exponent right over here is negative, if 1 minus p is less than zero, well think about it; then it's going to be one over M to some positive exponent, is one way to think about it. So as M approaches infinity, this whole thing is going to approach zero.

So this is actually going to be a situation where we converge, where we get to a finite value. So we add p to both sides; we have one is less than p; we converge. So there you have it; we have established this integral is going to converge only in the situation where p is greater than one.

If p is greater than one, you are going to converge, and if zero is less than p is less than or equal to one, you are going to diverge. Those are then the exact conditions because this p series converges if and only if this integral converges. These exact same constraints apply to our original p series.

Our original p series converges only in the situation where p is greater than one; then we converge, and if zero is less than p is less than or equal to one, we diverge. There you go.