Convergence of asymptotic series for $z^k/(z)_k$, $z\to\infty$

Question

Let $z>0$, $k\in\Bbb Z$, and $(s)_n=\Gamma(s+n)/\Gamma(s)$ denote the Pochhammer symbol. According to DLMF 5.11.13 as $z\to\infty$: $$ \frac{z^k}{(z)_k}\sim\sum_{\ell=0}^\infty\binom{-k}{\ell}B_\ell^{(1-k)}\frac{1}{z^\ell}, $$ where $B_n^{(k)}$ is the Norlund polynomial.

What conditions guarantee this asymptotic series converges?

What I have so far:

If $k=0,-1,-2,\dots$ then the series terminates after $1-k$ terms yielding $$ \frac{z^k}{(z)_k}=\sum_{\ell=0}^{-k}\binom{-k}{\ell}B_\ell^{(1-k)}\frac{1}{z^\ell}, $$ which in this case converges for all $z>0$. Furthermore, if $k=1$ then $B_\ell^{(0)}=\delta_\ell$ so that $$ \frac{1}{(z)_0}=1. $$ Also, of $k=2$ then $B_\ell^{(-1)}=\frac{1}{\ell+1}$ so that $$ \frac{z^2}{(z)_2}\sim\sum_{\ell=0}^\infty\binom{-2}{\ell}\frac{1}{\ell+1}\frac{1}{z^\ell}=\frac{1}{1+\frac{1}{z}}, $$ which obviously converges for $z>1$.

So what I still need is the case for $k=3,4,5,\dots$

Maybe one could use induction and noting that $$ \frac{z^{k+1}}{(z)_{k+1}}=\frac{1}{1+\frac{k}{z}}\frac{z^k}{(z)_k}? $$

Answer 1

Assume that $k \geq 2$ is an integer. You can write your expansion in terms of the Stirling numbers of the second kind as follows: $$ \frac{{z^k }}{{(z)_k }} \sim \sum\limits_{\ell = 0}^\infty {( - 1)^\ell S(\ell + k - 1,k - 1)\frac{1}{{z^\ell }}} . $$ From the asymptotics $$ S(\ell + k - 1,k - 1) \sim \frac{{(k - 1)^{\ell + k - 1} }}{{(k - 1)!}}, \quad \ell \to +\infty $$ we can see that the series converges if $k-1<|z|$ (you have to study the case $|z|=k-1$ separately). Alternatively, you can notice that the series is the Laurent expansion of the meromorphic function on the left-hand side. The function has simple poles at $z=-1,-2,\ldots,1-k$, whence the Laurent expansion will converge for $k-1<|z|$.

Answer 2

Piggybacking off @Gary's answer we also have the following proof by induction. There may be a couple refinements needed but this should work.

Setting $k=2$ we find $$ \frac{z^2}{(z)_2}\sim\sum_{\ell=0}^\infty\binom{-2}{\ell}\frac{1}{\ell+1}\frac{1}{z^\ell}=\frac{1}{1+\frac{1}{z}}, $$ which converges absolutely for $z>2-1$. Now assume the series in question converges absolutely for $z>k-1$. It follows that $$ \frac{z^{k+1}}{(z)_{k+1}}=\frac{1}{1+\frac{k}{z}}\frac{z^k}{(z)_k}=\sum_{\ell=0}^\infty(-k)^\ell \frac{1}{z^\ell} \sum_{\ell=0}^\infty (-1)^\ell S(\ell+k-1,k-1)\frac{1}{z^\ell}. $$ The first series converge if $z>k$ while the second converges absolutely if $z>k-1$. It follows that the product of these series must also converge if $z>k$. Furthermore, $$ \begin{aligned} \sum_{\ell=0}^\infty(-k)^\ell \frac{1}{z^\ell} \sum_{\ell=0}^\infty (-1)^\ell S(\ell+k-1,k-1)\frac{1}{z^\ell} &=\sum_{\ell=0}^\infty\sum_{m=0}^\ell(-k)^{\ell-m} (-1)^m S(m+k-1,k-1)\frac{1}{z^\ell}\\ &=\sum_{\ell=0}^\infty(-1)^\ell S(m+k,k)\frac{1}{z^\ell}, \end{aligned} $$ where we have used this identity. But this is our asymptotic series for $k+1$; hence, $$ \frac{z^{k}}{(z)_{k}}=\sum_{\ell=0}^\infty (-1)^\ell S(\ell+k-1,k-1)\frac{1}{z^\ell} $$ converges for $z>k-1$.