Convergence of a sum to $e^{-1/2}$? I have this sum:
$\displaystyle \sum _{k=0}^{\frac{n}{2}-m} \frac{\left(-\frac{1}{2}\right)^k}{k!} \exp \left(\frac{3 (m+k)-2 (m+k)^2}{n}\right)$.
I have some suspicion that it converges to $e ^{-1/2}$ for all $ m=1,2....$ when $n \rightarrow \infty$; but failed to prove that. Any help?
 A: The statement is true.
Let $S_{-1} = 0$ and $S_n$ be the partial sum $\displaystyle\;\;\sum_{k=0}^{n}\frac{1}{k!}(-\frac12)^k\;\;$ for $n \ge 0$.
Let $\alpha_{n,k}$ be the double sequence defined by
$$\alpha_{n,k} = \begin{cases}
\exp\left(\frac{3(m+k)-2(m+k)^2}{n}\right), & 0 \le k \le \frac{n}{2} - m\\
\\
0, & k > \frac{n}{2} - m.
\end{cases}$$
For each $n$, the $n^{th}$ entry of the sequence we are interested in can be rewritten as
$$
\sum _{k=0}^{\frac{n}{2}-m} \frac{1}{k!}(-\frac12)^k \alpha_{n,k} =
\sum_{k=0}^{\infty} (S_k - S_{k-1})\alpha_{n,k} = \sum_{k=0}^{\infty} S_k (\alpha_{n,k} - \alpha_{n,k+1})$$
As long as $m > \frac14$, it is easy to check $a_{n,k}$ is a non-negative decreasing sequence in $k$ for fixed $n$.
We have:


*

*For fixed $n$, 
$\displaystyle\quad\sum_{k=0}^{\infty} |\alpha_{n,k} - \alpha_{n,k+1}| = \alpha_{n,0} = 
e^{(3m-2m^2)/n} \le \max( e^{3m-2m^2}, 1 ) < \infty.$

*$\displaystyle\lim_{n\to\infty} \sum_{k=0}^{\infty} (\alpha_{n,k} - \alpha_{n,k+1})
   =\lim_{n\to\infty} a_{n,0} = \lim_{n\to\infty} e^{(3m-2m^2)/n} = 1.$

*For fixed $k$, $\displaystyle\quad\lim_{n\to\infty} (\alpha_{n,k}-\alpha_{n,k+1}) = 1 - 1 = 0.$
The double sequence $\alpha_{n,k}-\alpha_{n,k+1}$ satisfies the conditions for
Silverman-Toeplitz theorem and hence
$$\lim_{n\to\infty} \sum_{k=0}^{\infty}S_k(\alpha_{n,k}-\alpha_{n,k+1}) = \lim_{k\to\infty} S_k = \sum_{k=0}^{\infty} \frac{1}{k!}(-\frac12)^k = e^{-1/2}$$
Notes
Using Silverman-Toeplitz is not the most efficient way to prove the statement. There is a theorem 

Dominated convergence theorem for series
Let $\beta_{n,k}$ be a double sequence and $b_k$, $d_k$ be two sequences such that
  
  
*
  
*$\displaystyle \lim_{n\to\infty} \beta_{n,k} = b_k$ exists for every $k$.
  
*For every $n$ and $k$, $|\beta_{n,k}| < d_k$ and $\displaystyle\;\sum_{k} d_k < \infty$.
  
  
  i.e. the rows of the double sequence $\beta_{n,k}$ are dominated by a single
  non-negative sequence $d_k$ which has finite sum. Then
$$\sum_{k} b_k = \sum_{k} \lim_{n\to\infty} \beta_{n,k} = \lim_{n\to\infty} \sum_{k}\beta_{n,k}$$

One can verify 
$$\begin{align}
\beta_{n,k} = & \frac{1}{k!}(-\frac12)^k \alpha_{n,k}\\
b_k = & \frac{1}{k!}(-\frac12)^k\\
d_k = & \frac{1}{k! 2^k} \max( e^{3m-2m^2}, 1 )
\end{align}
$$
satisfies above condition for "DCT for series" and the statement follows immediately.
I can't find any wiki page for this "DCT for series" but here is a proof I find online.
