Computing radius of convergence for series without coefficients explicitly defined Consider the series
$$
1 - x^2 + x^4 - x^6 + \cdots = \sum_{n=0}^{\infty} a_n (x-3)^n.
$$
I'm trying to find the radius of convergence for this series. However, I'm having trouble coming up with a formula for the coefficients $a_n$ since the series is not centered at $x=0$.
At first glance, I thought that $a_n = 0$ for odd $n$ and that it alternated between $1$ and $-1$ for even $n$. However, computing the first several terms, I see that this is not the case.

Let $x=3+z$, then the LHS is $$\frac1{1+x^2}=\frac1{1+(z+3)^2}=\frac1{(z+3+i)(z+3-i)}=\frac{i/2}{z+3+i}-\frac{i/2}{z+3-i}.$$ One can deduce the series expansion in powers of $z$ but, if only the radius of convergence is concerned, the coefficients $(a_n)$ of the expansion are not needed: note simply that the disk of convergence is defined by the inequality $$|z|\lt\min\{|-3-i|,|-3+i|\}=\sqrt{10},$$ thus, the radius of convergence of the RHS is $R=\sqrt{10}$.
