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.
Best Answer
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).