Separation of variables and choosing solutions

Question

On Griffiths E&M page 145 he does an example of an uncharged metal sphere of radius R in a uniform E-field $E=E_0\hat{z}$. Now with the general solution $$V(r,\theta)=\sum_{l=0}^{\infty}\left(A_lr^l+\frac{B_l}{r^{l+1}}\right)P_l(cos(\theta))$$ and noting that $V=-E_0z+C$ And V=0 at r=R we get $$A_l R^l+\frac{B_l}{r^{l+1}}=0$$ This is fine. But then it says that for the solution as r$\rightarrow \infty$, our $B_l$ Becomes negligible and therefore we only deal with the $A_l$ terms. This is breaking my head a little bit because I thought that we always chose the non-problematic term as the solution i.e non-diverging or blowing up at zero. But the $A_l r^l$ terms will diverge as r$\rightarrow \infty$. What is going on here? Any help would be greatly appreciated!

Answer 1

You've written it out yourself: "$V= -E_0z+ C$". $z= r\cos\theta$, so $V\sim -E_0 r$ as $r\rightarrow \infty$. Then we do have to deal with the $A_l$ terms, but it doesn't mean all of them. We want to match the coefficients of $A_l$ to $V$, which means $A_1 \neq 0$ and $A_l = 0$ for $l\neq1$.

Answer 2

A point of clarification: As long as the series is converging, it is a well defined function.

For example, $e^{-x}=\sum_{n=0}^\infty \frac{(-x)^n}{n!} $

Here $A_n=\frac{(-1)^n}{n!}$ and the function $\rightarrow0$ as $x\rightarrow \infty$