Proving that the Legendre differential equation has solution of degree $n$ when $p=n$

Question

For $p\geq 0$ the Legendre differential equation is $$(1-t^2)y''-2ty'+p(p+1)y=0.$$ Two linearly independent solutions that I have found for this diff equation are $$y_1(t)=1+\sum_{n=1}^\infty(-1)^n\frac{[(p-2n+2)...(p-2)p][(p+1)(p+3)...(p+2n-1)]}{(2n)!}t^{2n}$$and $$t+\sum_{n=1}^\infty(-1)^n\frac{[(p-2n+1)...(p-3)(p-1)][(p+2)(p+4)...(p+2n)]}{(2n+1)!}t^{2n+1}.$$ How do I show that for $p=m$ (with $m$ an integer) the Legendre diff equation has a polynomial solution of degree $m$?

Answer 1

See my post About the Legendre differential equation for details of why the equation has a polynomial solution $P_n(x)$ of degree n (the p in your notation). You can find the other solution there also, for which the expansion does not terminate and which is normally denoted as $Q_n(x)$.

By the way the closed form you included in the question is not really practical in my opinion; better remember Rodrigues' formula $$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$ which clearly is a polynomial of degree n.