Worked example: interval of convergence | Series | AP Calculus BC | Khan Academy

So we have an infinite series here, and the goal of this video is to try to figure out the interval of convergence for this series. That’s another way of saying, for what x values, what range of x values is this series going to converge? And like always, pause this video and see if you can figure it out.

So when you look at this series, it doesn't fit cleanly into something like a geometric series or an alternating series. When I see something like this, I think about the ratio test because it tends to be pretty general. To apply the ratio test, we want to think about the limit—the limit as n approaches infinity of the n plus one term divided by the nth term and the absolute value of that. If this thing is less than one, then we are going to converge, and the x values that make this thing greater than 1, we are going to diverge. The x values that make this equal to 1, well then we are going to be inconclusive, and so we're gonna have to use other techniques to think about whether we're going to converge or diverge.

So let's just think about this. Let's just evaluate this. So let's do that limit as n approaches infinity of the absolute value of a sub n plus 1. Well, that's going to be x to the n plus 1. Let me color code this just so we know what we're doing. So this thing right over here is going to be x to the n plus 1 over n plus 1 times 5 to the n plus 1. We're going to divide that by the nth term. We're going to divide that by the nth term, and that's just going to be x to the n over n times 5 to the n. We're going to take the absolute value of this whole thing.

Now, let me simplify this. I'll simplify it down here. This is the same thing as x to the n plus one over n plus one times 5 to the n plus one times the reciprocal of this. So it's going to be n times 5 to the n over x to the n. We could simplify this. This is going to be equal to, let's see, you divide the numerator and denominator by x to the n, you're left with just an x. Then divide the numerator and denominator by 5 to the n, that is going to be—that's a 1. This is a 1, and then this is just going to be 5 to the n plus 1 divided by 5 to the n, that's just going to be a 5.

And so what do we have? We have x times n over distribute the five: 5n plus 5. Let me be careful there. Let me distribute the five, 5n plus 5. 5 times n, 5 times 1. 5n plus 1. 5, 5n plus 5.

All right, so let me just rewrite that. This is going to be equal to the limit as n approaches infinity of the absolute value of this thing. And actually, to help us with this limit, let me rewrite it a little bit. Let me divide the numerator and the denominator both by n. I'm not changing the value; I'm doing the same thing to the numerator and the denominator. I'm dividing it by the same value. So if I divide the numerator and denominator by n, this is going to be the same thing as x over 5 plus 5 to the n.

So when I divide the numerator and denominator by n, it becomes very clear what happens as n approaches infinity. As n approaches infinity, x doesn't change, 5 doesn't change, but 5 over n goes to 0. And so this limit is going to be equal to x over 5. So that's a pretty neat, clean thing.

So now we can use this to think about—actually, let me write this. This is going to be the absolute value of x over 5. And so now we can think about under which conditions is the absolute value of x over 5 going to be less than one and we're going to definitely converge? Under what conditions are we going to be greater than one and definitely diverge? And then on what conditions is it inconclusive?

So let's just see when we know we can converge. The absolute value of x over 5 is less than 1. This is our convergence situation. Well, that's the same thing as saying that negative 1 is less than x over 5, which is less than 1. If you multiply all the sides by 5, this is the same thing as negative 5 is less than x, which is less than 5.

So if we know that this is true, this is definitely going to be part of the interval of convergence. We know that if x meets these constraints, then our series is going to converge. But we're not done yet; we have to think about the inconclusive case.

So let's think about the scenario where the absolute value of x over 5 is equal to 1. Another way of thinking about this is this means that x over 5 is equal to 1, or x over 5 is equal to negative 1. This means that x is equal to 5 or x is equal to negative 5.

So these are the two inclusive cases using the ratio test. Let's test them out individually by looking back at the series and just substituting x equals 5 or x equals negative 5.

So in the first scenario, let me find a new color here. Let me use red. So the first scenario of x equals 5. Let’s go to our series, then the series is going to be the sum from n equals 1 to infinity of 5 to the n over n times 5 to the n. Well, this is just going to be equal to the sum n equals 1 to infinity of 1 over n, and this is a harmonic series. This is the p series where p is equal to one, and we know for p series, if p is equal to one, that's going to diverge. So this, and we know the harmonic series, we've done in other videos, this definitely diverges.

So this diverges, and you can do that by p series convergence test. If the p for a p series is one, well, you're going to diverge. Now, let's think about—so five is definitely not part of our interval of convergence.

Now, let's consider x equals negative five. When x equals negative 5, let me get another color going here. When x is equal to negative 5, then this thing is going to be equal to the sum from n equals 1 to infinity of negative 5 to the n. Actually, let me just write that as—I’ll write it out—negative 5 to the n over n times 5 to the n.

This is the same thing as the sum from n equals 1 to infinity of—we could write this as negative 1 to the n times 5 to the n over n times 5 to the n. And now this thing—this is an alternating harmonic series. And so you could actually use the—we—you might already know that that converges, or you could use the alternating series test.

And the alternating series test—it might be a little bit clearer if I write it like this—that this is an alternating series. So in an alternating series test, if we see that this thing is monotonically decreasing and the limit as n approaches infinity is zero, this thing converges. The alternating harmonic series actually converges, so this converges.

So given that this converges, you could view this as this boundary here. We would include that in our interval of convergence. So x doesn't just have to be strictly greater than negative 5; it could be greater than or equal to negative 5, but it has to be less than 5. This is our true interval of convergence.