$\sum_{n=0}^{\infty} c_{2n}$ and $\sum_{n=0}^{\infty} c_{2n+1}$ given $c_{n+2} = \frac{n(n+1)-\lambda}{(n+2)(n+1)}$

Legendre equation is

$$(1-x^2) y'' - 2xy' + \lambda y=0$$

We are interested in finding solutions in the range $$[-1,1]$$. We seek solutions around the ordinary point $$x=0$$ $$\sum_{n=0}^\infty c_n x^n$$ Since legendre equation has regular singular points at $$-1,1$$ , this series is guaranteed to converge in $$|x|<1$$. Substituting this power series into the differential equation one arrives at the following recursive relation

$$c_{n+2} = \frac{n(n+1)-\lambda}{(n+2)(n+1)}$$

Putting $$c_0 = 1$$ and $$c_1 = 0$$ gives the solution $$y_1(x)$$ Putting $$c_0 = 0$$ and $$c_1 = 1$$ gives the solution $$y_2(x)$$

Now if $$\lambda = l(l+1)$$ with $$l$$ a non-negative integer it can be shown that a solution to legendre equation is the legendre polynomial $$P_l(x) = \frac{1}{2^l l!} \frac{d^l}{dx^l}(x^2-1)^l$$ which is analytic everywhere.

However, if $$\lambda$$ is not of this form, the series solutions converges at the end points $$x=-1,1$$ ? If not, why not?