My first question is whether I can solve the Legendre equation for $l=0$, i.e. $$ (1-x^2)y''(x)-2xy'(x)+y(x)=0, $$ using the regular power series method around $x=0$. My second question is, even if you can solve the Legendre equation around $x=0$, how would you solve it around a regular singular point?
Say I want solve the equation around $x=1$, I would then use $$ y(x)=\sum_{n=r}^\infty{}a_n(x-1)^n $$ (where $a_r\neq0$) which when substituted in the Legendre equation would turn it into $$ (1-x^2)\sum_{n=r}^\infty{}a_nn(n-1)(x-1)^{n-2}-2x\sum_{n=r}^\infty{}a_nn(x-1)^{n-1}+\sum_{n=r}^\infty{}a_n(x-1)^n=0 $$ which after some manipulation looks like $$ \sum_{n=r}^\infty{}a_nn(n-1)(x-1)^{n-2}-\sum_{n=r}^\infty{}a_nn(n-1)(x-1)^{n}-2\sum_{n=r}^\infty{}a_nn(x-1)^{n}+\sum_{n=r}^\infty{}a_n(x-1)^n=0 $$ and after factoring all that can be factored $$ \sum_{n=r}^\infty{}a_nn(n-1)(x-1)^{n-2}+\sum_{n=r}^\infty{}[n(n-1)-2n+1]a_n(x-1)^n=0. $$
But how do I get the indicial equation out of this? I am supposed to find $r=0$ and $r=3$. But how?
I'll advance where you stopped. Replacing $n$ by $n+r$ and $n+r-2$ in the first and second sums respectively you get
$$ \sum_{n=0}^\infty a_{n+r}(n+r)((n+r)-1)(x-1)^{n+r-2}$$
$$+\sum_{n=2}^\infty{}[(n+r-2)((n+r-2)-1)-2(n+r-2)+1]a_{n+r-2}(x-1)^{n+r-2}=0 .$$
Simplify the above and you should be able to find advance.