Find the radius of convergence and interval of convergence of the series. ∑n=2^∞(-1)^n (x^n)/(4^

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

Key Concepts

Alternating Series Test

The alternating series test is used to determine the convergence of series whose terms alternate in sign. It states that an alternating series converges if the absolute values of the terms decrease monotonically to zero. This test is particularly useful when the convergence is conditional, meaning the series converges due to the cancellation of terms rather than their absolute summability.

Interval of Convergence

The interval of convergence is the set of all x-values for which the power series converges. After determining the radius of convergence, the endpoints of the interval need to be checked individually because the series behavior at these boundary points can vary; the series might converge absolutely, converge conditionally, or diverge at each endpoint.

Radius of Convergence

The radius of convergence is a crucial concept that defines the distance from the center of the power series within which the series converges absolutely. It is often computed using methods like the Ratio Test or the Root Test by finding the limit of |a?|^(1/n) toward the nth term, leading to a condition that |x - c| must be less than a certain number for convergence.

Power Series

A power series is an infinite series typically written as ? a? (x - c)?, where the coefficients a? and the variable x determine the convergence behavior. Its analysis involves determining the range of x (or values of x relative to the center c) for which the series converges, either absolutely or conditionally.

Transcript

00:01 Here we're given the summation of the minus 1, bow of the n, x bount n divided by 4 bount n times n, from 2d infinity.

00:11 In this question here, we need to apply the ratio test here.

00:18 So we have a ratio test, and then we need to compute the limit of the am plus 1 over an and the infinity.

00:30 And then we should get now the limit of now for the am plus 1 we should get the x bell am plus 1 over the 4 about n plus 1 times a l and the n plus 1 and now we divide by a n so we should get the multiplied by the reciprocal so we should get now will be the 4 b0 x power n times an of n divided by x power n here we see we can cancel the x -bel -n with this power, far -belled and with this power here, and then we can bring the x and the 4 here outside the limit.

01:10 Therefore we should get now equal to the absolute x over 4 times the limit of ln of n over the ln of m -1.

01:22 And as we see that as n goes to infinity, this limit here will equal to 1.

01:28 Therefore, we should get equal to absolute x over 4.

01:33 And now, to be convergence, this limit here must be smaller than 1.

01:38 And this isn't even only if absolute of x is smaller than 4, even only if absolute of x will be between minus 4 and 4 here.

01:50 And now for the first case, for the left end point here, x equal to minus 4, we get the series now will become the summation minus 1 power n and minus 4 power n divided by 4 power n from choose infinity so we cancel out everything here and then we have the submission one over l of n here and we know that this one will be diversioned because by the comparison test here one over l and is bigger than one over n and we know that that the summation of the 1 over n is diversion by the harmonic series here...

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