If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# 2015 AP Calculus BC 6a

## Video transcript

the Maclaurin series for a function f is given by and they give it an 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 if we have an infinite series so we go from N equals 1 to infinity and each term is a sub n the ratio test ratio test says all right let's consider the ratio between successive terms so we could say the ratio of a sub n plus 1 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 you know adding those terms as we're getting close to infinity so we'd want to take the limit as n approaches infinity here and if this limit exists let's say it's L and if L if L is less than 1 then the series the series converges converges so what we're gonna want to do and if L is greater than 1 it diverges it's equal to 1 it's 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 1 so let's do that so let's first think about this ratio so a sub n plus 1 over a sub n is going to be equal to so if we put n plus 1 into this expression here we're gonna have so let me make this clear so I'm gonna do a sub n plus 1 up here so we're gonna have negative 3 and I could write to the n plus 1 instead of an N and then minus 1 so it would be n plus 1 minus 1 well plus 1 minus 1 is just 0 so that's just the same thing as negative 3 to the N times X to the n plus 1 X to the n plus 1 over n plus 1 and plus 1 so that's a sub n plus 1 there and a sub n well that's just negative 3 to the n minus 1 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 thats the same thing as multiplying by the reciprocal of all of this stuff down here so it's negative negative 3 to the N X to the n plus 1 over n plus 1 times the reciprocal of this business so times n over over negative 3 to the n minus 1 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 1 this divided by X to the N is going to be X or X to the first power and we can divide the numerator in the denominator by negative 3 to the N minus 1 well this is just going to be 1 and if you divide negative 3 to the N by negative 3 to the N minus 1 well that's just going to be negative 3 to the first power so this is all going to be negative 3 X n over over n over n plus 1 so now let's think about what the limit of the 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 3 X n over n plus 1 now some of you might recognize if we focus on end we have the same degree up here same degree down here this both end 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 what 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 let me let me write it 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 at 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 can do as long if I multiply or divide the denominator by this the numerator and the denominator by the same thing I'm not changing it value so if I do if I divide both of them by n and the numerator I just get one in the denominator and divided by n is 1 1 divided by n is plus one over n and so this might be a little bit clearer that ok as n approaches infinity well we don't know this doesn't deal with n but this over here 1 over N is going to approach 0 and so this whole thing is going to approach 1 and so the limit is going to be 3 times the absolute value of x and so remember this series converges is if this limit if this limit is less than 1 so converges converges converges if 3 times the absolute value of x is less than 1 or we could say the absolute value of x just divide both sides by 3 is less than 1/3 and so we have just found our radius of convergence R so we could say R is equal to 1/3 this Maclaurin series is going to converge as long as the absolute value of x is less than 1/3 or we could say our radius of convergence is equal to the our radius of convergence is equal to 1/3 so there you go
AP® is a registered trademark of the College Board, which has not reviewed this resource.