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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.

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

Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 8

Power Series

Sequences

Series

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

01:59

In mathematics, a series is, informally speaking, the sum of the terms of an infinite sequence. The sum of a finite sequence of real numbers is called a finite series. The sum of an infinite sequence of real numbers may or may not have a well-defined sum, and may or may not be equal to the limit of the sequence, if it exists. The study of the sums of infinite sequences is a major area in mathematics known as analysis.

02:28

In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed. Like a set, it contains members (also called elements, or terms). The number of elements (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. Formally, a sequence can be defined as a function whose domain is either the set of the natural numbers (for infinite sequences) or the set of the first "n" natural numbers (for a finite sequence). A sequence can be thought of as a list of elements with a particular order. Sequences are useful in a number of mathematical disciplines for studying functions, spaces, and other mathematical structures using the convergence properties of sequences. In particular, sequences are the basis for series, which are important in differential equations and analysis. Sequences are also of interest in their own right and can be studied as patterns or puzzles, such as in the study of prime numbers.

02:01

Suppose that the power ser…

02:40

Show that if $ \lim_{n \to…

03:19

Prove that if the power se…

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?

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