2015 AP Calculus BC 6a

- [Voiceover] The Maclaurin series for a function f is given by, they give it to us in sigma notation and then they expand it out for us, and converges to f of x for the absolute value of x being less than R, where R is the radius of convergence of the Maclaurin series. Part a. Use the ratio test to find R. So first of all, if terms like Maclaurin series and radius of convergence or even convergence or ratio test seem foreign to you or you have some foggy memories of it, you might want to review all of those concepts on Khan Academy. We actually have multiple videos and exercises on each of these concepts on Khan Academy. But if you kind of know what it is, I will give you a little bit of a reminder for the ratio test. So the ratio test tells us if we have an infinite series, so we go from n equals one to infinity, and each term is a sub n, the ratio test, ratio test, says alright, let's consider the ratio between successive terms. So we could say the ratio of a sub n plus one over a sub n and in particular we want to focus on the absolute value of this ratio. And this by itself, it might not be a constant like we would see in a geometric series, it actually might be a function of n itself. And so we want to see the behavior of this ratio as n gets really, really, really large, as we're kind of adding those terms as we're getting close to infinity. So we want to take the limit as n approaches infinity here. And if this limit exists, let's say it's L, and if L is less than one, then the series, the series converges. So what we're gonna want to do, and if L is greater than one, it diverges, if it's equal to one, it's inconclusive. And so what we want to do here is we want to figure out the absolute value of the ratio here, take the limit and then see for what x values does that limit, will that limit be less than one? So let's do that. So let's first think about this ratio. So a sub n plus one over a sub n is going to be equal to... So if we put n plus one into this expression here, we're gonna have... so let me make this clear, So I'm gonna do a sub n plus one up here, so we're gonna have negative three, and I could write to the n plus one instead of an n, and then minus one, so it would be n plus one minus one. Well plus one minus one is just zero so that's just the same thing as negative three to the n, times x to the n plus one, x to the n plus one, over n plus one. So that's a sub n plus one there. And a sub n, well that's just negative three to the n minus one times x to the n over n. So what does this right over here simplify to? This is going to be equal to, well we could just say this divided by that's the same thing as multiplying by the reciprocal of all of this stuff down here. So it's negative three to the n, x to the n plus one over n plus one, times the reciprocal of this business. So times n over negative three to the n minus one, x to the n. And so can we simplify this? Well we can divide both the numerator and the denominator here by x to the n. So this divided by x to the n is just one. This divided by x to the n is going to be x, or x to the first power. And we can divide the numerator and the denominator by negative three to the n minus one. Well this is just going to be one. And if you divide negative three to the n by negative three to the n minus one, well that's just going to be negative three to the first power. So this is all going to be negative three x n over n plus one. So now let's think about what the limit of the absolute value of this as n approaches infinity is. So the limit the limit as n approaches infinity of the absolute value of negative three x n over n plus one. Now some of you might recognize if we focus on n, we have the same degree up here, same degree down here, it's both n to the first, so these are going to, the n over n plus one is going to approach one. And so you might say "Okay, well this is going to be the absolute value of negative three x", but if you want to make that a little bit clearer this is equal to the limit as n approaches infinity of, and I'll write it this way, so I could write the absolute value of negative three x or the absolute value of negative three x is the same thing as three times the absolute value of x times the absolute value of, if I divide the numerator here by n, I would get one and if I divide the denominator by n, I could do... if I multiply or divide the numerator and the denominator by the same thing, I'm not changing the value. So if I divide both of them by n, in the numerator I just get one, in the denominator, n divided by n is one, one divided by n is plus one over n. And so this might be a little bit clearer that as n approaches infinity, well we don't know, this doesn't deal with n. But this over here, one over n is going to approach zero, and so this whole thing is going to approach one and so the limit is going to be three times the absolute value of x. And so remember, this series converges if this limit is less than one. So converges, converges, converges if three times the absolute value of x is less than one. Or we could say the absolute value of x, just divide both sides by three, is less than one third. And so we have just found our radius of convergence. So we could say R is equal to one third. This Maclaurin series is going to converge as long as the absolute value of x is less than one third. Or we could say our radius of convergence is equal to, our radius of convergence is equal to one third. So there you go.