TITLE: Help evaluating a limit
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: Let $X_n$ denote a Poisson random variable with parameter $\alpha\sqrt{n}$ and $Y_n$ a negative binomial random variable with parameters $(n,\frac12)$. Recall that this means that, for every $k\geqslant0$,
$$
P[X_n=k]=\mathrm e^{-\alpha\sqrt{n}}\frac{(\alpha\sqrt{n})^k}{k!},\qquad P[Y_n=k]=2^{-n-k} {n-1+k\choose k}.
$$
Thus, assuming that $X_n$ and $Y_n$ are independent, the $n$th sum you are considering is
$$
S_n=P[X_n+Y_n\leqslant n-1].
$$
First, $Y_n$ can be realized as the sum of $n$ i.i.d. random variables with geometric distribution of parameter $\frac12$, in particular, the central limit theorem states that $Y_n=n+\sqrt{2n}Z_n$ where $Z_n$ converges in distribution to a standard random variable $Z$. Second, the variance of a Poisson random variable being equal to its mean, $X_n=\alpha\sqrt{n}+\sqrt{n}T_n$ where $T_n\to0$ in distribution. Hence,
$$
\frac{X_n+Y_n-n+1}{\sqrt{n}}=\alpha+T_n+\sqrt{2}Z_n+\frac1{\sqrt{n}}
$$
converges in distribution to $\alpha+\sqrt{2}Z$.
Thus,
$$
S_n=P\left[\frac{X_n+Y_n-n+1}{\sqrt{n}}\leqslant0\right]\to S=P[\alpha+\sqrt{2}Z\leqslant0],
$$
and
$$
S=1-\Phi\left(\frac{\alpha}{\sqrt2}\right)=\frac12\left(1-\mathrm{erf}\left(\frac{\alpha}{2}\right)\right).
$$