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?

Note that $(-1)^n = (-1)^{n-2}$. Hence, $$\sum_{n=0}^\infty(-1)^n\frac{4^{n-2}(x-2)}{(n-2)!} = (x-2) \sum_{n=0}^\infty\frac{(-4)^{n-2}}{(n-2)!} = (x-2) \left(\sum_{n=0}^\infty\frac{(-4)^{n}}{n!} \right)$$
where we have interpreted $\dfrac1{(-1)!} = 0 = \dfrac1{(-2)!}$.
This is a reasonable interpretation since $\dfrac1{\Gamma(0)} = 0 = \dfrac1{\Gamma(-1)}$.
Now recall that $$\sum_{n=0}^\infty\frac{y^{n}}{n!} = \exp(y).$$
Can you now conclude that the series converges no matter what value $x$ takes?
