Simplifying a summation involving factorials

Question

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.

Answer 1

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.

Answer 2

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.