Domain limitations on generating function for Legendre polynomials The generating function for legendre polynomials is $$\frac{1}{\sqrt(1-2ut+u^2)}=\sum_{i=0}^\infty u^iP_i(t)$$ where $P_i$ is $i^{th}$ legendre polynomial however for binomial expansion of left hand side with exponent  -$\frac12$, we need to have $$|(u^2-2ut)|\le1$$. So how do we prove equality of two sides when this condition doesn't hold ? Or is it really a domain limitation ?
 A: We consider the sequence $\{P_n(t)\}$ of Legendre polynomials. We describe how to construct a generating function
\begin{align*}
G(t,u)=\sum_{n=0}^\infty P_n(t)u^n
\end{align*}
and how to derive the region of convergence. We closely follow example 7.4 from  Asymptotics and Special Functions by F.J.W. Olver.

We  recall  Rodrigues'  formula
\begin{align*}
P_n(t)=\frac{(-1)^n}{2^nn!}\frac{d^n}{dt^n}\left\{\left(1-t^2\right)^n\right\}
\end{align*}
  and get  using  Cauchy's  integral formula  for the $n$-th derivative of an analytic function Schläfli's integral
\begin{align*}
P_n(t)=\frac{1}{2^{n+1}\pi i}\int_{\mathcal{C}}\frac{(z^2-1)^n}{(z-t)^{n+1}}dz
\end{align*}
in which $\mathcal{C}$ is any simple closed contour that encircles $z=t$; here  $t$ may be real or complex.  For fixed  $\mathcal{C}$  and sufficiently  small $|u|$,  the series
  \begin{align*}
\sum_{n=0}^\infty\frac{(z^2-1)^nu^n}{2^{n+1}\pi  i(z-t)^{n+1}}
\end{align*}
  converges uniformly   with  respect  to  $z\in\mathcal{C}$, by  the  M-test.  By integration and summation we obtain
  \begin{align*}
\frac{1}{2\pi   i}\int_{\mathcal{C}}\left\{1-\frac{(z^2-1)u}{2(z-t)}\right\}^{-1}\frac{dz}{z-t}=\sum_{n=0}^\infty  P_n(t)u^n=G(t,u).
\end{align*}
  It follows
  \begin{align*}
G(t,u)=-\frac{1}{\pi     i}\int_{\mathcal{C}}\frac{dz}{uz^2-2z+(2t-u)}=-\frac{1}{u\pi  i}\int_{\mathcal{C}}\frac{dz}{(z-z_1)(z-z_2)}
\end{align*}
  where
  \begin{align*}
z_1=\frac{1-\sqrt{1-2tu+u^2}}{u},\qquad   z_2=\frac{1+\sqrt{1-2tu+u^2}}{u},
\end{align*}

We   observe if $u\to 0$,  then   $z_1\to t$ and $|z_2|\to\infty$.  Hence  for sufficiently small  $|u|$, $\mathcal{C}$ contains $z_1$ but not $z_2$. The residue theorem yields
\begin{align*}
G(t,u)=-\frac{2}{u}\frac{1}{z_1-z_2}=\frac{1}{\sqrt{1-2tu+u^2}}
\end{align*}
We  conclude, the desired expansion is given by 
\begin{align*}
\color{blue}{\frac{1}{\sqrt{1-2tu+u^2}}=\sum_{n=0}^\infty P_n(t)u^n}\tag{1}
\end{align*}
provided that $|u|$ is sufficiently small and the chosen branch of the square root tends to $1$ as $u\to 0$.
For $t\in[-1,1]$ the singularities of the left-hand side of (1) both lie on the circle $|u|=1$, hence in this case the radius of convergence of the series on the right-hand side is unity.
