Find the radius of convergence and interval of convergence of the series. ∑n=1^∞ 2^n n^2 x^n

Name: Problem 10, Chapter 11: Calculus
Uploaded: 2021-07-20T19:33:39-08:00
Duration: 8 min 17 s
Channel: Tamara Worner
Description: Find the radius of convergence and interval of convergence of the series. ∑n=1^∞ 2^n n^2 x^n

Find the radius of convergence and interval of convergence...

Key Concepts

Ratio Test

The ratio test is a method used to determine the convergence of infinite series by examining the limit of the absolute value of the ratio of consecutive terms. When applied to power series, this test facilitates finding the radius of convergence by establishing when the series satisfies the convergence criteria.

Power Series

A power series is an infinite series of the form ? a? (x - c)?, where the coefficients a? and the variable x determine the behavior of the series near the center c. Understanding power series is crucial for analyzing the convergence of functions represented as infinite sums.

Radius of Convergence

The radius of convergence is a non-negative value that defines the interval within which a power series converges absolutely. It is typically determined by applying convergence tests, such as the ratio or root test, to the series coefficients to identify the boundary for convergence.

Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges. It includes the open interval defined by the radius of convergence and may also involve testing the endpoints separately, since convergence behavior can differ at these boundary points.

Transcript

00:01 Okay, the key to this problem again is we want to find the radius of convergence and the interval of convergence.

00:07 And to do that, we need to look at the limit of the ratio test.

00:13 So the next term divided by the previous term and figure out when that limit is less than one.

00:19 That'll give us our radius and then that will tell us our interval and then we check the end points individually.

00:26 So for this example, we're going to use this series n equals 1 to infinity of 2 to the n times n squared times x to the n.

00:43 And we want to find out which x is can i plug in to make this series converge.

00:50 So we're going to start with step one.

00:53 We're going to calculate the limit as n goes to infinity.

00:59 Okay, the top of our fraction is just the terms in the series with an n plus 1 instead of an n.

01:07 So 2 to the n plus 1, n plus 1 squared, x to the n plus 1.

01:17 Divided by the bottom of our fraction is the nth term of the series.

01:22 So 2 to the n, n squared, x to the n.

01:26 All right.

01:28 Now, what usually happens in these problems is there's a lot of simplifying.

01:33 So here, 2 to the n plus 1 divided by 2 to the n.

01:38 Again, remember, subtract your exponents.

01:41 We're going to get just a 2 there.

01:43 And likewise, with the x to the n and the x to the n plus 1 will be left with just an x in the numerator.

01:50 So i can simplify this, and i'll get the limit as n goes to infinity of the absolute value of 2x in my numerator.

02:02 And then the n plus 1 and the n are both squared so i can rewrite that like this n plus 1 over n quantity squared.

02:14 My 2x here is constant with respect to the n in the limit.

02:20 That means i can factor it out of the limit.

02:23 So it's in absolute values.

02:25 So let me pull out the absolute value of 2x times the limit as n goes to infinity of the absolute value.

02:35 Of n plus 1 squared, n plus 1 over n squared.

02:42 Super.

02:43 Now, again, we should recognize this limit is equal to 1.

02:48 If you need to convince yourself of that, try lopetal's rule for calculating limits with n's instead of x's, but that limit is just going to be 1.

02:58 All right.

03:00 Now, what i need to have is i need that to be less than 1.

03:05 So i need just the absolute value of 2x.

03:10 To be less than one.

03:13 Okay, so to do that, that's just the same as two times the absolute value of x is less than one.

03:19 So the absolute value of x is less than one half.

03:26 And so that one half, that is my radius of convergence.

03:39 So my radius of convergence is equal to one half.

03:47 And i'm thinking my interval is negative a half, is less than x is less than a half.

03:54 But i need to check the end points...