The generating function of the product of Legendre polynomials for the same $n$ is given by


\begin{aligned}
\sum\_{n=0}^{\infty} z^{n} \mathrm{P}\_{n}(\cos \alpha) \mathrm{P}\_{n}(\cos \beta)&=\frac{\mathrm{F}\left(\frac{1}{2} ,\frac{1}{2} ;1;\frac{4z\sin \alpha \sin \beta }{1-2z\cos (\alpha +\beta )+z^{2}}\right)}{\sqrt{1-2z\cos (\alpha +\beta )+z^{2}}} \\&=\frac{2K\left(\sqrt{\frac{4z\sin \alpha \sin \beta }{1-2z\cos (\alpha +\beta )+z^{2}}}\right)}{\pi \sqrt{1-2z\cos (\alpha +\beta )+z^{2}}}
\end{aligned}
where $P\_n$ is a Legendre polynomial, $F$ is a hypergeometric function and $K$ is a complete elliptic integral of the first kind.
<https://www.researchgate.net/publication/269015726_A_generating_function_for_the_product_of_two_Legendre_polynomials>


However, in the study of quantum physics, we need the similar result as above for following expression:
\begin{align}
\sum\_{n=0}^{\infty} z^{n} \mathrm{P}\_{n}(x) \mathrm{P}\_{n-1}(x)
\end{align}


Is there any result known? Any hints for this problem would be appreciated.



Generating function of the product of Legendre polynomials
Using the recursion relation
$$
P\_{n-1} (x) = x P\_n (x) - \frac{x^2 -1}{n} \frac{d}{dx} P\_n (x) \ ,
$$
you can reduce your expression to a sum of the generating function you quote and a combined derivative (the $d/dx$) and integral w.r.t. $z$ (to generate the $1/n$) thereof.


