<s>[INST] How do I formally show the radius of convergence of the Taylor series of $f(x)=x^6 - x^4 + 2$ at $a=-2$? This is an exercise in Stewart's Calculus (Exercise 19, Section 11.10 Taylor and Maclaurin Series):

Find the Taylor series for $f(x)$ centered at the given value of a. [Assume that f has a power series expansion. Do not show that $R_n(x) \to 0.$ Also find the associated radius of convergence.
Here $f(x)=x^6 - x^4 + 2$ and $a=-2$.


I'm having trouble finding a general formula of this Taylor series and therefore, also having problems finding the radius of convergence since I can't perform the ratio test.
Here is what I know:
\begin{align}
f'(x) = 6x^5 - 4x^3,\quad     &f''(x) = 30x^4 - 12x^2,\\
f'''(x) = 120x^3 - 24x,\quad  &f^{(4)}(x) = 360x^2 - 24,\\
f^{(5)}(x) = 720x,\quad         &f^{(6)}(x) = 720.
\end{align}
And at $a=-2$,
\begin{align}
f(-2) = 50,\quad     &f'(-2) = -160,\\
f''(-2) = 432,\quad  &f'''(-2) = -912,\\
f^{(4)}(-2) = 1416,\quad         &f^{(5)}(-2) = -1440,\\
f^{(6)}(-2) = 720.\quad    &
\end{align}
I'm having trouble finding the general formula for each term. Without it, how am I supposed to find the radius of convergence?

Added:
So the general term I have for the n-th derivative of $f$ is:
$$f^{(n)}(x) = \frac{6!x^{6-n}}{(6-n)!}$$
So far the general term I have for the Taylor Series is:
$$\sum_{n=0}^{\infty} \frac{6! 2^{6-n}}{(6-n)!n!}(x+2)^n$$
I can see why the radius of convergence is $\infty$: because for any $x$the series converges.
But how do I show this formally? Can I use the ratio test?
 [/INST] The Taylor series around $a$ is simply given by $$\begin{align}f(a+x)&=(a+x)^6-(a+x)^4+2\\&=(a^6-a^4+2)+(6a^5-4a^3)x+(15a^4-6a^2)x^2+(20a^3-4a)x^3+(15a^2-1)x^4+6ax^5+x^6\end{align}$$
or
$$f(x)= (a^6-a^4+2)+(6a^5-4a^3)(x-2)+(15a^4-6a^2)(x-a)^2+(20a^3-4a)(x-a)^3+(15a^2-1)(x-a)^4+6a(x-a)^5+(x-a)^6$$
and, as almost all coefficients are $=0$, converges for all $x$.
</s>