💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Suppose that the power series $\sum c_n (x - a)^n$ satisfies $c_n \not= 0$ for all $n.$ Show that if $\lim_{n \to \infty} \mid c_n/c_{n + 1} \mid$ exists, then it is equal to the radius of convergence of the power series.

## Go through the usual procedure of the Ratio Test and you end up with$R=\lim _{n \rightarrow \infty}\left|\frac{c_{n}}{c_{n+1}}\right|$

Sequences

Series

### Discussion

You must be signed in to discuss.
##### Top Calculus 2 / BC Educators  ##### Catherine R.

Missouri State University ##### Samuel H.

University of Nottingham Lectures

Join Bootcamp

### Video Transcript

okay for this problem because we're already using a here. Will you be in ter for tow? These terms you know, the ratio test says that limit as n goes to infinity. Absolute value of P N plus one over being. If we get convergence, then this should be less than one. And if it's equal to one, it could converge. It might not converge. So we have X minus A to the n. So well, this right this outs we have seen C n plus one here and then X minus a to the end, plus one. And then we're dividing by CNN Times X minus a to the end. So there's cancellations that'LL happen here X minus a the n plus one divided by X minus a The end is just going to turn into x minus a. I can remember when we're trying to figure out the radius of convergence. We looked to see when this is less than one, when it's less than one. We we know that we for sure have convergence. So now we can divide both sides by the absolute value of X minus a. And once we do that, we get limit as n goes to infinity Absolute value of C N plus one over seeing he's less than one over absolute value of X minus a Okay, And now if we one eye, take the inverse of both sides and we have to switch this inequality symbol here. So this is taking the inverse of both sides. We get limited in goes to infinity of C n o ver si n plus one. And when we're taking the inverse again, this has toe flip. We get that we get that this is greater than absolute value of X minus. So we get the X minus is trapped between this value which will just call it is called C for now. So we have that minus c is less than X minus. A is less than C and now we can add eight of both sides to get a minus. C is less than acts is less than a plus c no. So we get convergence when X is between a minus C and a plus C. So the length of our interval of convergence is a plus C minus a minus e, which is so we get a ADA cancelled and we have C minus minus C. So this is to see that the length of our interval of convergence and then our radius of convergence is going to be half of that. So, indeed we get their radius of convergence is C? #### Topics

Sequences

Series

##### Top Calculus 2 / BC Educators  ##### Catherine R.

Missouri State University ##### Samuel H.

University of Nottingham Lectures

Join Bootcamp