From limn →∞ √(|(-1)^n((1+2 i)/(2))^n|)=limn →∞|(1+2 i)/(2)|=(√(5))/(2) we see that the radius o

From limn →∞ √(|(-1)^n((1+2 i)/(2))^n|)=limn →∞|(1+2...

From $\lim _{n \rightarrow \infty} \sqrt[n]{\left|(-1)^{n}\left(\frac{1+2 i}{2}\right)^{n}\right|}=\lim _{n \rightarrow \infty}\left|\frac{1+2 i}{2}\right|=\frac{\sqrt{5}}{2}$ we see that the radius of convergence is $R=2 / \sqrt{5}$. The circle of convergence is $|z+2 i|=2 / \sqrt{5}$.

Key Concepts

Circle of Convergence

The circle of convergence is the set of all complex numbers that satisfy |z - c| < R, where c is the center of the series and R is the radius of convergence. This concept is key in understanding the domain over which a power series converges in the complex plane.

Complex Modulus

The complex modulus measures the magnitude or absolute value of a complex number, calculated as the square root of the sum of the squares of its real and imaginary parts. It is essential for evaluating convergence when the terms of a series are complex numbers.

Nth Root Test

The nth root test involves taking the limit of the nth root of the absolute value of the coefficients of a series. In complex analysis, this test connects directly to the radius of convergence by identifying the boundary at which the power series transitions from convergent to divergent.

Radius of Convergence

The radius of convergence determines the distance from the center of a power series within which the series converges. It is a fundamental concept in the study of power series, especially in complex analysis, and can be calculated using tests such as the nth root test or the Cauchy-Hadamard formula.

Transcript

00:01 Here we're given the series of the form to x minus 1 power of the n divided by 5 and then times 5 power n and then we have this one is it will be a little bit confusing here so i guess is n here and from 1 to infinity and we need to use the ratio test here and in the ratio test we need to compute the limit of the am plus 1 over a .n.

00:36 N goes to infinity.

00:38 And then we should get now, for the i .m.

00:42 This 1, we get the 2x minus 1.

00:47 And now about n plus 1, dividing by 5 about n plus 1.

00:53 And then we have the square root of the n plus 1.

00:56 And now we divide by the i .n, so we should have the times with the reciprocal.

01:01 So we'll be 5 about n.

01:02 The square root of the n over the 2x minus 1 power of the n.

01:08 Here we see we can cancel the 2x minus 1 power and will dispel on the top, 5 power and will dispel in the bottom, and then we can put the 2 of them into the same square root.

01:20 And the last thing we can do will be repringition outside, even the 5 here as well.

01:26 As a result, we get equal to the absolute 2x minus 1 and then dividing by 5.

01:36 Times the limit of the square root of the n over n plus 1, and goes to infinity.

01:46 Here we can push the limit inside the square root, and then we have the absolute of 2x minus 1, divided by 5 times the square root on the limit, n over n plus 1.

02:00 And this limit here clearly equals to 1...

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