# Choosing $A_l=0$ to guarantee bounded potential in infinity

I'm taking a course in Electrodynamics and quite often, when using the spherical approach $$\Phi=\sum\limits_{l~=~0}^{\infty}\left(A_lr^l+B_lr^{-(l+1)}\right)P_l(\cos\gamma),$$ there's the argument that, e.g. when calculating the potential outside a sphere, the $A_l$ must be $0$ to guarantee boundedness of the potential at infinity. But if e.g. I consider $A_l=\frac{(-1)^l}{l!}$, giving me $e^{-r}$, the boundedness would still be given. Where is my mistake, or is there another reason for choosing $A_l=0$?

It would only equal $e^{-r}$ for $\gamma=0$. For $\gamma = \pi$ you get $P_l(-1) = (-1)^l$, changing your answer to $e^r$ in that direction.