# Legendre Polynomial manipulations

Given Legendre polynomial generating function $$\sum_{n=0}^\infty P_n (x) t^n = \frac{1}{\sqrt{(1-2xt+t^2)}}$$ Show that $$P_n (1)=1$$ and $$P_n (-1)=(-1)^n$$

Substituting $x=1$ in the given equation yields
$$\sum_{n=0}^\infty P_n (1) t^n = \frac{1}{(1-2t+t^2)^{\frac12}} = \frac{1}{((1-t)^2)^{\frac12}} = = \sum_{n=0}^{\infty} t^n$$
which gives $P_n(1)=1$. You can do the same with the other one.
substitute $x=1$: $$\sum_{n=0}^\infty P_n (1) t^n = \frac{1}{(1-2t+t^2)^{\frac12}} = \frac{1}{((1-t)^2)^{\frac12}} = \frac{1}{1-t}$$ and can you expand that in powers of $t$?
Similar method for $x=-1$.