How can I prove differential recurrence relation of Legendre polynomials without generating function?

Question

What I am trying to prove is $P'_{n+1}(x)−P'_{n−1}(x)=(2n+1)P_n(x)$

What I can use are here:

1. $P_0(x)=1$, $P_1(x)=x$
2. $\int_{-1}^1 P_m(x)P_n(x) \;dx = 0 \;(n\neq m)$ (the orthogonality of Legendre polynomial)
3. $(n + 1)P_{n+1}(x) = (2n + 1)xP_n(x)-nP_{n-1}(x)$

By differentiating 3, I can get $(n+1)P'_{n+1}(x) = (2n+1)P_n(x)+(2n+1)xP'_n(x)-nP'_{n-1}(x)$

So what I should prove is $nP_n(x)=xP'_n(x)-P'_{n-1}(x)$.

How to prove this without using generating function of Legendre polynomial?

Answer 1

Another possibility is the following one: show that $P_{n+1}'-P_{n-1}'$ (which is a polynomial with degree $n$) is orthogonal to each Legendre polynomial $P_k$ with the only exception of $P_n$, by computing $\int_{-1}^{1}(P_{n+1}'(x)-P_{n-1}'(x))P_k(x)\,dx $ through integration by parts.