Ordinary generating function for Lah's numbers $L(n,k)$

Question

I'm studing signed Lah's numbers $L(n,k) = (-1)^n\frac{n!}{k!}\binom{n-1}{k-1}$.

It's easy to show that exponential generating function for the sequence $\{L(n,k)\}_{n\geq0}$ is equal to $$ \sum\limits_{n=0}^{\infty}L(n,k)\frac{t^n}{n!} = \frac{(-1)^k}{k!}\left(\frac{t}{1+t}\right)^k. $$ So what's about ordinary generating function $\sum\limits_{n=0}^{\infty}L(n,k)x^n = ?$. I'm trying to apply connection between ordinary generating function $A(x)=\sum a_n x^n$ and exponential generating function $B(t) = \sum a_n \frac{t^n}{n!}$: $$ A(x) = \int\limits_0^{\infty}e^{-t}B(xt)dt, $$ to prove that $\sum\limits_{n=0}^{\infty}L(n,k)x^n = x\prod\limits_{k=1}^{\infty}\frac{x+k}{1-kx}$.

So how to prove it?

Answer 1

For fixed $k$, note that $L(n,1) = (-1)^n n!$, so for $k=1$ the power series would be $$ \sum\limits_{n} (-1)^n n! x^n. $$ The problem is, this power series doesn't converge for any real $x \neq 0$, so it can't be expressed as an analytical function. The terms grow even faster for larger $k$, so it's true for them as well.

If you want an OGF with a fixed $n$ rather than fixed $k$, you essentially get Laguerre polynomials, which are also quite far from the form that you want to prove in your question.

As I also mentioned in the comments, the RHS of your expression doesn't meaningfully converge, and if we limit the product up to $k$ rather than up to $\infty$, we get something that seems to have nothing to do with Lahs numbers. So, I don't think this question can be meaningfully answered without further clarification.

Answer 2

The ordinary generating function is a divergent series for all $k\ge 1$. However, it is a Borel summable one. The Borel sum may be expressed in terms of the Kummer $U$-function: \begin{align*} A(x) &= \frac{{( - 1)^k }}{{k!}}\int_0^{ + \infty } {{\rm e}^{ - t} \left( {\frac{{xt}}{{1 + xt}}} \right)^k {\rm d}t} = \frac{{( - 1)^k }}{{k!}}\frac{1}{x}\int_0^{ + \infty } {{\rm e}^{ - t/x} \left( {\frac{t}{{1 + t}}} \right)^k {\rm d}t} \\ & = ( - 1)^k \frac{1}{x}U(k + 1,2,1/x) = ( - 1)^k U(k,0,1/x) \end{align*} (cf. $(13.4.4)$ and $(13.2.40)$). The formal generating function arises as the asymptotic expansion as $x\to 0^+$ of this function: $$ ( - 1)^k U(k,0,1/x) \sim x^k \sum\limits_{n = 0}^\infty {( - 1)^{n + k} \frac{{(k)_n (k + 1)_n }}{{n!}}x^n } = \sum\limits_{n = k}^\infty {L(n,k)x^n } $$ (use Watson's lemma or $(13.7.3)$).