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 $2n-1$?

Question

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.

Answer 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.

Answer 2

$\newcommand{\bi}{\binom}$Let us prove the more general conjecture suggested by T. Amdeberhan.

Let \begin{equation*} \begin{aligned} &s_n(x,y) \\ &:=\sum_{k=0}^\infty\frac{n!}{(k+n)!}x^kL_n^k(x)L_n^k(y) \\ &=\sum_{k=0}^\infty\frac{n!}{(k+n)!} \sum_{j=0}^n \frac{(-1)^j}{j!}\bi{n+k}{n-j}y^j \, \sum_{i=0}^n \frac{(-1)^i}{i!}\bi{n+k}{n-i}x^{k+i} \\ &=n! \sum_{p=0}^\infty \sum_{j=0}^n y^j \, x^{p-j} \sum_{i=0}^n 1(i+j\le p)\frac{(-1)^{i+j}}{i!j!} \frac1{(p-i-j+n)!}\\ &\qquad\qquad\qquad\qquad\quad\times\bi{n+p-i-j}{n-i}\bi{n+p-i-j}{n-j}. \end{aligned} \tag{1}\label{1} \end{equation*} So, for natural $n$, after some algebra we get \begin{equation*} \frac{s_n(x,y)-s_{n-1}(x,y)}{(n-1)!} =\sum_{p=0}^\infty \sum_{j=0}^n c_{n,p,j}\, y^j x^{p-j}, \tag{2}\label{2} \end{equation*} where \begin{equation*} c_{n,p,j}:=\sum_{i=0}^n 1(i+j\le p)F(j,i) \tag{3}\label{3} \end{equation*} and \begin{equation*} F(j,i):=F_{n,p}(j,i):=\frac{(-1)^{i+j} (n p-i j) (n+p-i-j-1)!}{i! j! (n-i)! (n-j)! (p-i)! (p-j)!}; \end{equation*} note that $c_{n,p,j}=0$ if $j>p$.

Following Peter Taylor's answer, let \begin{equation*} R(j,i):=\frac{i \left(n+p+n p-i-j-i j\right)}{(j+1) \left(n p-i j\right)} \end{equation*} and then \begin{equation*} G(j,i):=R(j,i)F(j,i). \end{equation*} Then we get the Wilf–Zeilberger identity \begin{equation*} F(j+1,i)-F(j,i)=G(j,i+1)-G(j,i). \tag{4}\label{4} \end{equation*} Take now any $p\ge2n$. Then the factor $1(i+j\le p)$ in \eqref{3} is $1$ for $i,j\le n$, and hence, in view of the telescoping property provided by \eqref{4}, $c_{n,p,j}$ does not depend on $j\in\{0,\dots,n\}$: \begin{equation*} \begin{aligned} c_{n,p,j}&=\sum_{i=0}^n F(j,i) \\ &=\sum_{i=0}^n F(n,i) =\frac{(-1)^n n}{n!n!(n+q)!}\sum_{i=0}^n (-1)^i \bi ni=0. \end{aligned} \end{equation*} So, $c_{n,p,j}=0$ for $p\ge2n$.

That is, by \eqref{2}, $s_n(x,y)-s_{n-1}(x,y)$ is a polynomial (in $x,y$) of degree $\le2n-1$. Also, $s_0(x,y)=e^x$. Thus, $s_n(x,y)-e^x$ is a polynomial of degree $\le2n-1$.

Substituting here $x$ for $y$, we get the conjecture in the OP.

Answer 3

Conjecture. It appears that this generalization holds: $$ \sum_{k=0}^{\infty}\frac{n!}{(k+n)!}x^kL_n^k(x)L_n^k(y)=e^x+P_{2n-1}(x,y), $$ where $P_{2n-1}(x,y)$ is a polynomial of degree $2n-1$.