Help evaluating a limit I have the following limit:
$$\lim_{n\rightarrow\infty}e^{-\alpha\sqrt{n}}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}\sum_{m=0}^{n-1-k}\frac{(\alpha\sqrt{n})^m}{m!}$$
where $\alpha>0$.
Evaluating this in Mathematica suggests that this converges, but I don't know how to evaluate it.  Any help would be appreciated.

I would start even more simple-mindedly by replacing the inner sum
with its infinite $n$ value of $e^{\alpha \sqrt{n}}$.
This cancels out the outer expression, so we are left with
$\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}$.
Doing some manipulation,
$\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}
= \sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose {n-1}}
= \sum_{k=n-1}^{2n-2}2^{-k-1} {{k}\choose {n-1}}
$.
As often happens, it is late and I am tired and not sure exectly what to do next,
so I'll leave it at this.
