Legendre Polynomial manipulations

Question

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

Not sure where to start with either

Answer 1

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.

Answer 2

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 $x=-1$.