Find the radius of convergence and interval of convergence of the series. ∑n = 2^∞ (x^2n)/(n(lnn

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

Key Concepts

Power Series

A power series is an infinite sum of the form ? a? (x - c)? where each term involves a coefficient and a power of x. In general, it represents a function as an infinite polynomial, and its convergence depends on the value of x relative to the center c.

Radius of Convergence

The radius of convergence is the distance from the center of the power series within which the series converges absolutely. It is typically calculated using convergence tests like the Ratio or Root Test by examining the limiting behavior of the coefficients.

Interval of Convergence

The interval of convergence is the set of all x-values for which a power series converges. It is determined by the radius of convergence and requires additional endpoint analysis to check for convergence or divergence at the boundary points.

Convergence Tests

Convergence tests, such as the Ratio Test and the Root Test, are methods for determining the absolute or conditional convergence of a power series. These tests analyze the behavior of the series' terms as n approaches infinity and are crucial for finding the radius of convergence and assessing convergence at the endpoints.

Transcript

00:01 Okay, for this problem, we'll use the ratio test to first just figure out where we get convergence.

00:08 So limit as n goes to infinity of a .n plus 1 over a .n.

00:12 And by a .n, we mean this whole chunk here, including the x values.

00:20 So this is x to the 2 times n plus 1 over n plus 1, ln of n plus 1 squared.

00:37 So that's just our a .n plus 1 term.

00:40 By a n which is multiplying by the reciprocal of a n so now we're multiplying by n times natural log of n squared and we're dividing by x to the 2n so x to the 2 times n plus 1 is the same thing as x to the 2n multiplied by x squared so x to the 2n will cancel out with this x to the 2n and we'll just have x squared there and then here we have an n here we have an n plus 1 so we'll group those terms together.

01:21 And then here, this is an exponent of two.

01:23 This is an exponent of two.

01:26 So we'll also lump these terms together.

01:33 As n goes to infinity, natural log of n divided by natural log of n plus 1 is going to go to 1.

01:38 You can see that by applying lopatatal's rule.

01:43 And as n goes to infinity, n over n plus 1 is also going to go to 1.

01:47 So this is just going to be absolute value of x squared and we want for that to be less than one okay but x squared is already going to be something that's non -negative so x squared the only real condition here is that x squared is less than one okay and this means that x is going to be between minus one and one so we'd get that by taking the square root of both sides square root and then doing the plus minus and you'd get that x is between minus square root of one and positive square root of one, which is just minus 1 and 1.

02:33 Okay, so our interval of convergence is somewhere from minus 1 to 1.

02:39 At this point, we don't know whether or not we include minus 1 and whether or not we include 1.

02:45 The length of this interval is going to be 2, though, which means that the radius of convergence is 2 divided by 2, which is 1.

02:53 So 1 is our radius of convergence...

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