Simplifying a summation involving factorials How does one show this?
$$
\exp(-x) \sum_{k=0}^\infty x^k \frac{(k+m)!}{(k!)^2} = L_m(-x) m!,
$$ where $m$ is a positive integer, and $L_{m}(x)$ is the $m$th order Laguerre polynomial.
 A: One possible direction: suffices to show that
$$
e^x L_m(-x)
 = \sum_{k=0}^\infty \frac{(k+m)!}{k! \cdot m!} \frac{x^k}{k!}
 = \sum_{k=0}^\infty \binom{k+m}{k} \frac{x^k}{k!},
$$
which can be interpreted as a generating function.
A: Perhaps this is a bit of a convoluted approach, but I have used this once before but with the Legendre polynomial.
$$\exp(-x) \sum_{k=0}^\infty x^k \frac{(k+m)!}{(k!)^2} = L_m(-x) m!$$
Substitute $x \to -x$
$$\sum_{k=0}^\infty (-1)^k x^k \frac{(k+m)!}{(k!)^2 m!} = L_m(x) e^{-x}$$
Multiply both sides by a dummy variable $t^m$ and sum from $m=0$ to $\infty$
$$\sum_{m=0}^\infty t^m\sum_{k=0}^\infty (-1)^k x^k \frac{(k+m)!}{(k!)^2 m!} = \sum_{m=0}^\infty L_m(x) e^{-x}t^m$$
RHS is simplified by the definition of the Laguerre polynomial (from the generating function)
$$\sum_{m=0}^\infty t^m\sum_{k=0}^\infty (-1)^k x^k \frac{(k+m)!}{(k!)^2 m!} = e^{-x}\frac{e^{-tx/(1-t)}}{1-t}=\frac{e^{x/(t-1)}}{1-t}$$
Reorder the summation of the LHS and expand the RHS with the taylor series for $e^x$
$$ \sum_{k=0}^\infty x^k\sum_{m=0}^\infty t^m(-1)^k  \frac{(k+m)!}{(k!)^2 m!} =\frac{e^{x/(t-1)}}{1-t}=\sum_{k=0}^\infty \frac{-1}{(t-1)^{k+1}k!}x^k$$
Compare coefficients to get:
$$\sum_{m=0}^\infty t^m(-1)^k  \frac{(k+m)!}{(k!)^2 m!} =\frac{-1}{(t-1)^{k+1}k!}$$
$$\sum_{m=0}^\infty t^m(-1)^k  \frac{(k+m)!}{k! m!} =\frac{-1}{(t-1)^{k+1}}$$
I believe this is a known identity, if not, I can add a simple proof by induction. The issue I have with this is that I have ignored all convergence issues.
