Recurrence relations for Legendre polynomials prove by power series

Question

Given that $(1-2tx+t^2)\dfrac{\partial G}{\partial x} - Gt = 0$ and the generating function $G(x;t) = \dfrac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=0}^{\infty}P_n(x)t^n,$

show that $$P'_{n+1} - 2xP'_n + P'_{n-1} = P_n$$ \

Since the power series, $G(x;t)=\sum_{n=0}^\infty P_n(x)t^n$ and $\frac{\partial G(x;t)}{\partial x} = \sum_{n=0}^\infty P'_n(x)t^n$, by substituting into $(1-2tx+t^2)\dfrac{\partial G}{\partial x} - Gt = 0$ one gets,

$$(1-2tx+t^2)\sum_{n=0}^\infty P'_n(x)t^n - t\sum_{n=0}^\infty P_n(x)t^n = 0 $$

$$\sum_{n=0}^\infty \left(P'_n(x)-2xP'_n(x)t^{n+1}+P'_n(x)t^{n+2}\right)-\sum_{n=0}^\infty P_n(x)t^{n+1}=0$$

$$\sum_{n=0}^\infty \left(P'_n(x)-2xP'_{n-1}(x) + P'_{n-2}(x)\right)t^n - \sum_{n=0}^\infty P_{n-1}(x)t^n = 0 $$

$$\sum_{n=0}^\infty \left(P'_n(x)-2xP'_{n-1}(x) + P'_{n-2}(x) - P_{n-1}(x)\right) = 0 $$

I have gotten the above steps, which I can't show to the desired result. where have I done incorrectly? Help appreciated.

Answer 1

$$\sum_{n=0}^\infty \left(P'_n(x)\color{red}{t^n}-2xP'_n(x)t^{n+1}+P'_n(x)t^{n+2}\right)-\sum_{n=0}^\infty P_n(x)t^{n+1}=0$$ $$\color{red}{P_0'(x) + t(P_1'(x) -2xP_0'(x))~+} \sum_{\color{red}{n=2}}^\infty \left(P'_n(x)-2xP'_{n-1}(x) + P'_{n-2}(x)\right)t^n - \sum_{\color{red}{n=1}}^\infty P_{n-1}(x)t^n = 0$$ $$\color{red}{P_0'(x) + t(P_1'(x) -2xP_0'(x) - P_0(x))~+} \sum_{\color{red}{n=2}}^\infty \left(P'_n(x)-2xP'_{n-1}(x) + P'_{n-2}(x)-P_{n-1}\right)\color{red}{t^n} = 0$$

This is a power series in $t$. The only way a convergent power series can be $0$ for all $t$ in its radius of convergence is if all of its coefficients are $0$ (as can be proven by repeatedly differentiating with respect to $t$). Therefore $$P_0'(x) = 0\\P_1'(x) - 2xP_0'(x) - P_0(x) = 0$$ and for all $n \ge 2$, $$P'_n(x)-2xP'_{n-1}(x) + P'_{n-2}(x)-P_{n-1} = 0$$ All that remains is another shift of $n$ (which could have been avoided by being more careful earlier).