Convergence of $\sum_{n=0}^\infty(-1)^n\frac{4^{n-2}(x-2)}{(n-2)!}$ What theorem should I use to show that $$\sum_{n=0}^\infty(-1)^n\frac{4^{n-2}(x-2)}{(n-2)!}$$ is convergent no matter what value $x$ takes?

Ratio (D'alembert's) test, as proposed by Integral:
$$a_n:=(-1)^n\frac{4^{n-2}}{(n-2)!}\Longrightarrow \left|\frac{a_{n+1}}{a_n}\right|=\frac{4^{n-1}}{(n-1)!}\frac{(n-2)!}{4^{n-2}}=\frac{4}{n-1}\xrightarrow [n\to\infty]{} 0$$
so that the convergence radius is $\,\infty\,$.
