Consider the generating function of Legendre polynomials:
$$\frac{1}{\sqrt{1 - 2xt + t^2}} = \sum\limits^{\infty}\_{n=0} P\_n(x)t^n$$
Is it true that for $0<x<1, t=1$ series of Legendre polynomials converges to the function on the left-hand side, i.e.


$$\frac{1}{\sqrt{2-2x}} \stackrel{?}{=} \sum\limits^{\infty}\_{n=0} P\_n(x)$$


I know that the radius of convergence of $\sum\limits^{\infty}\_{n=0} P\_n(x)t^n$ equals to one so we need to figure out behaviour on the boundary of the disk of convergence.



Convergence of the series of Legendre polynomials
The answer is yes. By a theorem of Fatou


Theorem [Fatou]
If $a\_n\to0$ and the function $f(z)=\sum\_{n=0}^\infty a\_nz^n$ is analytic at 
the point $z=1$, then the series $\sum\_{n=0}^\infty a\_n$ converges with value 
$f(1)$. 


we only have to show that $\lim\_{n\to\infty}P\_n(x)=0$ for $0<x<1$. To see this we note the asymptotic expansion
$$P\_n(\cos\theta)\asymp \sqrt{\frac{2}{\pi n\sin\theta}}\,\sin\bigl((n+\tfrac12)\theta+\tfrac\pi4\bigr),\qquad n\to\infty,\quad \delta\le\theta\le \pi-\delta.$$


Fatou's theorem is found in Korevaar book (p.~148) on Tauberian Theory, the asymptotic expansion of Legendre Polynomials in the book of Lebedev on Special functions eq. (4.6.7). 


