I have the following expression:
$$
\sum\_{k=0}^{\infty}\frac{n!}{(k+n)!}x^k(L\_n^k(x))^2,
$$
where
$$
L\_n^k(x)=\sum\_{j=0}^n(-1)^j\binom{n+k}{n-j}\frac{x^j}{j!}
$$
is the usual associated Laguerre polynomial and $k\in\mathbb N$. In particular $\int\_0^{\infty}e^{-x}x^k(L^k\_n(x))^2=\frac{(k+n)!}{n!}$.


I am trying to figure out a way to simplify this sum. I am not an expert on special functions, and I would appreciate some references or hints. I have been trying quite hard using a few recurrence relations and other formulas found in the literature.


What I would like to prove is the following:
$$
\sum\_{k=0}^{\infty}\frac{n!}{(k+n)!}x^k(L\_n^k(x))^2=e^x+P\_{2n-1}(x),
$$
where $P\_{2n-1}(x)$ is a polynomial of degree $2n-1$ (if $n=0$, we set $P\_{-1}=0)$. This is clearly true when $n=0$, and one can easily prove for small values of $n$ ($n=1,2,...$). An alternative way is to differentiate $2n$ times and prove that the resulting sum gives $e^x$, but this approach remains complicated (at least, for me).


I don't know if this is something known, I would appreciate a reference in that case.



Infinite sum of Laguerre polynomials: is $\sum_{k=0}^{\infty}\frac{n!}{(k+n)!}x^kL_n^k(x)^2=e^x+P(x)$, with $P$ a polynomial of degree $2n-1$?
Everything becomes simpler if add some parameters and start the sum at $k=-n$ instead of $k=0$. Note that if $k$ is a negative integer with $-n\le k \le -1$ then $L\_n^k(x)$ is a polynomial in $x$ divisible by $x^{-k}$.


Then we have the formula
$$\sum\_{k=-n}^\infty \frac{n!}{(k+n)!}z^k L\_n^k(x)L\_n^k(y)=e^z\sum\_{i=0}^n\frac{1}{i!}\binom{n}{i}\left(\frac{(x-z)(y-z)}{z}\right)^i,$$
which can be proved by expanding the right side and simplifying, using Vandermonde's theorem.
Thus
\begin{multline\*}
\sum\_{k=0}^\infty \frac{n!}{(k+n)!}z^k L\_n^k(x)L\_n^k(y)\\
= e^z\sum\_{i=0}^n\frac{1}{i!}\binom{n}{i}\left(\frac{(x-z)(y-z)}{z}\right)^i - \sum\_{k=-n}^{-1} \frac{n!}{(k+n)!}z^k L\_n^k(x)L\_n^k(y).
\end{multline\*}
If we set $z=x$ the terms in the first sum on the right with $i>0$ vanish and we get
$$
\sum\_{k=0}^\infty \frac{n!}{(k+n)!}x^k L\_n^k(x)L\_n^k(y)
=e^x - \sum\_{k=-n}^{-1} \frac{n!}{(k+n)!}x^k L\_n^k(x)L\_n^k(y),
$$
so this gives an explicit formula for the OP's polynomials.


