The description of the problem goes:
Let the sequence $a_0, a_1,...$ be defined by the equation:
$$1-x^2+x^4-x^6+... = \sum_{n=0}^\infty a_n(x-3)^n,\;0<x<1.$$
Find $\limsup_{n\to \infty}(\lvert a_n\rvert ^\frac{1}{n}).$
This problem has shown on my Calculus test and haunted me for a long time. At first glance it seems like to find the coefficients of Taylor series of $f(x )=\frac{1}{1+x^2}$ centered at $x_0=3$. However the coefficients $a_n=\frac{f^\left (n\right)(3)}{n!}$ is not easy at all to compute. I've done some work and found it can be solved through some methods in complex analysis, which I do not have now. I was wondering whether it can be solved using Calculus ?
Note: I have another problem which I believe can be solved by similar argument:
$$\sum_{n=0}^\infty (-1)^n\left(\frac{x}{3}\right)^n=\sum_{n=0}^\infty a_n(x+1)^n,\;-3<x<1.$$
Find $\lim_{n\to \infty}(\lvert a_n\rvert ^\frac{1}{n}).$
Many thanks to anyone helps in advance!