I am sorry if this question isn't clear, I couldn't think of a better way to phrase it. I am a Physics student trying to solve the angular component of the wave function for a particle in a central potential. I am sure that most if not all of you are familiar with the problem. Here is what I get for the polar wave function.

$$ \frac{\sin(\theta)}{\Theta(\theta)}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta}\right) + A\sin^2(\theta) - m^2 = 0 $$

This is simply the differential equation for the Associated Legendre polynomial $P^m_l$ specifically where $A=l(l+1)$ for some integer, $l$ after a change of coordinates $x \rightarrow \cos(\theta)$. The original Associated Legendre differential equation is shown below.

$$ \left(1-x^2\right) \frac{d^2 P_l^m(x)}{dx^2} - 2x \frac{d P_l^m(x)}{dx} + \left(l(l+1)-\frac{m^2}{(1-x^2)}\right)P_l^m(x) = 0 $$

My problem is that in all references I have read on the subject it isn't clear why $A$ must equal $l(l+1)$. I understand that the series solution for the polar equation when transformed to the $x$ variable, $\Theta(\theta) \rightarrow y(x)$ yields the important recurrence relation:

$$ \left(1-x^2\right) \frac{d^2 y(x)}{dx^2} - 2x \frac{d y(x)}{dx} + \left(A-\frac{m^2}{(1-x^2)}\right)y(x) = 0 \\ y(x) = \left(1-x^2\right)^{m/2} f(x) \\ \left(1-x^2\right) \frac{d^2f(x)}{dx^2} -2(m+1)x \frac{df(x)}{dx} + \left(A - m(m+1)\right)f(x) = 0 \\ f(x) = \sum_{k=0}^\infty f_k x^k \\ f_{n+2} = \frac{(n+m)(n+m+1)-A}{(n+1)(n+2)}f_n $$ And specifying that $A=l(l+1)$ for some positive integer, $l$ cuts off higher order components of $x$ which generates the associated Legendre polynomial; however, $x=\cos(\theta)$, $0\leq\theta\leq\pi$ so $-1\leq x\leq1$. I cannot find any proof that demonstrates why this series diverges for all $x$ in this range if higher order $x$ terms are not cut off by $A$. $$ f(x) = \sum_{k=0}^\infty f_k x^k \rightarrow \mathrm{diverge/converge?} $$ Moreover the the actual solution to $\Theta(\theta)$ is $\left(1-x^2\right)^{\frac{m}{2}} f(x)$ due to the earlier substitution so I need to determine whether this diverges or converges for all $-1\leq x\leq 1$ $$ \Theta(\theta) = (1-x^2)^{\frac{m}{2}}\sum_{k=0}^\infty f_k x^k \rightarrow \mathrm{diverge/converge?} $$ Thank you for reading this far and I hope I have explained my problem adequately, I have tried to be concise as I can. I got the impression from a similar question that the answer might have something to with Hilbert space so I have been reading about that topic but I would appreciate whatever pointers I could get in the right direction.

I am sorry if this question isn't clear, I couldn't think of a better way to phrase it. I am a Physics student trying to solve the angular component of the wave function for a particle in a central potential. I am sure that most if not all of you are familiar with the problem. Here is what I get for the polar wave function.

$$ \frac{\sin(\theta)}{\Theta(\theta)}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial \Theta(\theta)}{\partial \theta}\right) + A\sin^2(\theta) - m^2 = 0 $$

This is simply the differential equation for the Associated Legendre polynomial $P^m_{\ell}$ specifically where $A=\ell(\ell+1)$ for some integer, $\ell$ after a change of coordinates $x \rightarrow \cos(\theta)$. The original Associated Legendre differential equation is shown below.

$$ \left(1-x^2\right) \frac{d^2 P_\ell^m(x)}{dx^2} - 2x \frac{d P_\ell^m(x)}{dx} + \left(\ell(\ell+1))-\frac{m^2}{(1-x^2)}\right)P_\ell^m(x) = 0 $$

My problem is that in all references I have read on the subject it isn't clear why $A$ must equal $\ell(\ell+1)$. I understand that the series solution for the polar equation when transformed to the $x$ variable, $\Theta(\theta) \rightarrow y(x)$ yields the important recurrence relation:

$$ \left(1-x^2\right) \frac{d^2 y(x)}{dx^2} - 2x \frac{d y(x)}{dx} + \left(A-\frac{m^2}{(1-x^2)}\right)y(x) = 0 \\ y(x) = \left(1-x^2\right)^{m/2} f(x) \\ \left(1-x^2\right) \frac{d^2f(x)}{dx^2} -2(m+1)x \frac{df(x)}{dx} + \left(A - m(m+1)\right)f(x) = 0 \\ f(x) = \sum_{k=0}^\infty f_k x^k \\ f_{n+2} = \frac{(n+m)(n+m+1)-A}{(n+1)(n+2)}f_n $$ And specifying that $A=\ell(\ell+1)$ for some positive integer, $l$ cuts off higher order components of $x$ which generates the associated Legendre polynomial; however, $x=\cos(\theta)$, $0\leq\theta\leq\pi$ so $-1\leq x\leq 1$. I cannot find any proof that demonstrates why this series diverges for all $x$ in this range if higher order $x$ terms are not cut off by $A$. $$ f(x) = \sum_{k=0}^\infty f_k x^k \rightarrow \mathrm{diverge/converge?} $$ Moreover the the actual solution to $\Theta(\theta)$ is $\left(1-x^2\right)^{\frac{m}{2}} f(x)$ due to the earlier substitution so I need to determine whether this diverges or converges for all $-1\leq x\leq 1$ $$ \Theta(\theta) = (1-x^2)^{\frac{m}{2}}\sum_{k=0}^\infty f_k x^k \rightarrow \mathrm{diverge/converge?} $$ Thank you for reading this far and I hope I have explained my problem adequately, I have tried to be concise as I can. I got the impression from a similar question that the answer might have something to with Hilbert space so I have been reading about that topic but I would appreciate whatever pointers I could get in the right direction.