Convergence of asymptotic series for $z^k/(z)_k$, $z\to\infty$ 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
$$
\frac{z^{k+1}}{(z)_{k+1}}=\frac{1}{1+\frac{k}{z}}\frac{z^k}{(z)_k}?
$$
 A: 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 $k-1<|z|$.
A: 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 $z>k-1$.
