We have the Legendre operator, $$ \mathcal{L}u = - \frac{d}{dx}\bigg[ (1-x^2)\frac{du}{dx}\bigg], $$ and we want to find its eigenvalues and eigenfunctions, therefore we must solve the following ODE, $$ (1-x^2) \frac{d^2u_n(x)}{dx^2} - 2x \frac{du_n(x)}{dx} + \lambda_nu_n(x) = 0 $$ We will solve it using the series method around $x=0$, therefore we suppose that $ u_n(x)=\sum_{i=0}^\infty a_ix^i$ and we obtain the following expression: $$ (2a_2+\lambda_na_0) + (6a_3+(\lambda_n-2)a_1)x+ \sum_{i=2}^\infty\big[(i+1)(i+2)a_{i+2}+(\lambda_n-i(i+1))a_i\big]x^i=0$$. Obtaining the followinf recurrence sequence,
- $2a_2+\lambda_na_0 = 0 \implies a_2=-\frac{a_0\lambda_n}{2}$
- $6a_3+(\lambda_n-2)a_1 = 0 \implies a3=-\frac{a_1(\lambda_n-2)}{6}$
- $(i+1)(i+2)a_{i+2}+(\lambda_n-i(i+1))a_i = 0 \implies a_{i+2}=-\frac{a_i(\lambda_n-i(i+2))}{(i+1)(i+2)}$
We can separete the original serie into 2 new series one with the even terms and the other with the odd terms. To study the convergence of this series we use the D'Alembert criterion (or ratio test) and we have that, $$ \lim_{i\to\infty}\frac{|a_{i+2}x^{i+2}|}{|a_{i}x^{i}|} = x^2 \cdot \lim_{i\to\infty}\frac{|(\lambda_n-i(i+1))|}{|(i+1)(i+2)|} = x^2 $$ Therefore it converges when $x\in(-1,-1)$.
Question
I have tried to study the convergence for $x=1$ and $x=-1$ and i know it converges for $\lambda_n=n(n+1)$ but i couldn't prove there are no other values of $\lambda_n$ for which the series is convergent. Any bibliography or some prove of it?
Thank you so much!