Evaluation of the limit for a power series yields an absolute value polynomial; how do I find R?

Question

I'm asked to find the radius of convergence for $\sum\frac{(x-1)^{2n}}{4^n}$

I used the root test, as this is easiest because of all the n-th power terms. However, when I evaluate the limit, I have an $|(x-1)^2|$ term, and this must be $<4$. If I expand the polynomial, I can move the $+1$ over to the right, such that I'm left with $|(x^2-2x)|$ < 3, and $|x(x-2)| < 3$, but we have not learned how to determine the center and Radius of convergence from such a form. I need it to be in the $|x-a| < R$ form.

Help is much appreciated, and I apologize for all the posts I am making regarding homework questions. I'm just really struggling, and my professor hasn't had time to help me.

Answer 1

Hint. One may observe that $$ |(x-1)^2|<4 $$ is equivalent to $$ |x-1|<2. $$

Answer 2

Other way, more general for more advanced examples: define

$$a_n=\begin{cases}\frac1{2^n},&n\;\text{is even}\\{}\\0,&n\;\text{is odd}\end{cases}$$

and now you really have the power series about $\;x_0=1\;$ :

$$\sum_{n=0}^\infty\frac{(x-1)^{2n}}{4^n}=\sum_{n=0}^\infty a_n(x-1)^n$$

use now the Cauchy-Hadamard formula (theorem) to find the convergence radius:

$$\frac1R=\overline{\lim}_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\sqrt[n]\frac1{2^n}=\frac12\implies R=2$$

so your power series converge s in $\;|x-1|<2\iff -2<x-1<2\iff-1<x<3\;$ . You can now directly check that both end poinst are not included in the convergence interval.