Interval of convergence for derivative and integral | Series | AP Calculus BC | Khan Academy

Times in our dealings with power series, we might want to take the derivative or we might want to integrate them. In general, we can do this term by term. What do I mean by that? Well, that means that the derivative of f prime of x is just going to be the derivative of each of these terms. So, that's going to be the sum from n = 1 to infinity, and let's see, the derivative of x to the N is n * x to the N minus 1.

So I could write this as n * x to the N minus 1 all of that over n. These n's will cancel out, so this is just going to be x to the N minus one. So, this is taking the derivative with respect to x. Similarly, we could integrate, and we could evaluate the integral of f of x dx. This is going to be equal to some constant plus if we integrate this term by term.

So this is going to be equal to the sum from n = 1 to infinity, and let's see, we increment the exponent so x to the n + 1, and then we divide by that. So, times n + 1 times this n right over here. This is a common technique that you will see when dealing with power series. We're going to go a little bit more into the details because you can only do this for x values within the interval of convergence for the power series.

As we will see, the interval of convergence for these different series is slightly different. The intervals are very similar, but what happens at the end point is different. So, I encourage you to pause this video and see if you can figure out the interval of convergence for each of these series. This is the integral of our original series and this is the derivative of our original series.

So, let's start with our original series; let's figure out the interval of convergence. We could do that using the ratio test. For the ratio test, we would want to do the limit as n approaches infinity of a sub n + 1, so that's going to be x to the n + 1 over n + 1 divided by a sub n. So, that's x to the N over N.

We want to take the absolute value of that; that's going to be the limit as n approaches infinity. Let's see, if you divide this and this by x to the n, that's going to be a one, and this is just going to be an x. And then this n is going to end up up top. So, this is going to be x to the n over n + 1. This is equal to the limit as n approaches infinity.

Uh, let's see, if we divide the numerator and the denominators here by one over if we divide both of the numerator and denominator by n, we're going to get x over 1 + 1/n. And what is this going to be? Well, this term is going to go to zero, so this is just going to be equal to the absolute value of x.

The ratio test tells us that this series is convergent if this right over here is less than one, it's divergent if this is greater than one, and it's inconclusive if this equals one. So we know, let's write that down: we know we are convergent for the absolute value of x less than one.

When this thing evaluates when it is less than one, we know that we are divergent when this thing is greater than one; when the absolute value of x is greater than one. But what about when the absolute value of x is equal to one? That's where the ratio test breaks down; we have to test that separately.

So, let's look at the scenario where x is equal to one. When x equals one, this series is the sum from n = 1 to infinity of 1 to the N / n. Well, that's just going to be 1/n; this is the harmonic series or the p series where our p is one, and we've seen in multiple videos that this diverges.

So when x equals 1, we diverge. But what about when x equals -1? When x equals -1, this thing becomes the sum from n = 1 to infinity of (-1) to the N / n, and this is often known as the alternating harmonic series. This one, by the alternating series test, actually converges, and we've seen that in multiple videos.

So it turns out the interval of convergence for our original thing right over here, our interval of convergence is where x can be greater than or equal to -1 or I could say -1 is less than or equal to x. Because if x is -1, we still converge, but then x has to be less than 1 because right at 1 we diverge, so we can't say less than or equal to.

So, this is the interval of convergence for our original function. What about the interval of convergence for this one right over here when we take the derivative? Well, when we take the derivative, this is the same thing as x to the 0 plus x to the 1 plus x to the 2, and we go on and on and on. Now you might recognize this; this is a geometric series with a common ratio of x.

Geometric series where our common ratio, often denoted by r, is equal to x. We know that a geometric series converges only in the situation where the absolute value of our common ratio is less than one. So, in this situation when we took the derivative for f prime of x, our interval of convergence is almost the same.

So here our interval of convergence is going to be x has to be between -1 and 1, but it can't be equal to -1 or 1; we would actually diverge at those points. So, notice these are almost the same; if we view these as series centered at zero, the radius of convergence is the same.

We can go one above, one below one above, one below, and that's in general true as we take derivatives and integrals. But the endpoints of our interval of convergence can be different. To continue to see this, I encourage you to use the ratio test to figure out what the interval of convergence is for this, for the antiderivative for the integral here.

What you will see is the radius of convergence is the same; we can go one above zero and one below zero. We have to be in that interval, but as you will see, this one converges for x = -1 or x = 1. I'll just cut to the chase here.

So, interval of convergence... let me write that in yellow. The interval of convergence for this top one converges for -1 < x ≤ 1. So notice they all have the same radius of convergence, but the interval of convergence differs at the endpoint. If you want to prove this one for yourself, I encourage you to use a very similar technique that we used for our original function.

Use the ratio test; you're going to come to this conclusion right over here, and then test the cases when x is equal to 1 and x is equal to -1. You will see when x is equal to 1, you have an alternating p series, so that's going to converge. Then when x equals -1, you're going to have a p series where the denominator has a degree larger than one or something similar to a p series, and you can establish that will also converge in that scenario as well.