Evaluating the limit of this complicated sum? So, I'm trying to evaluate this limit:
$$\lim_{m \to \infty} \sum_{k=0}^{m}\sum_{j=0}^{m-k} \frac{(-1)^j(x-a)^j}{k! j! (\frac{1}{m})^{j+k}}$$
As you can see, it is quite complicated, and evaluating it paper-and-pencil style seems like a daunting task. So, I tried to use a calculator, namely, WolframAlpha, but even through Wolfram, the computation time is exceeded, and I don't have pro, so I can't use extra computation time. Also, I'm not at all savvy with computers, so I can't use a programmed calculator unless I spend weeks learning how to program, which I would much rather not do. Other than Wolfram, I don't know of any other calculators I could use.  
So, can I get a little help here? Perhaps doing it by pencil and paper is much less rigorous than I thought, but I'm only really interested in an answer, if the limit even does exist and the sum converges. 
 A: Let's do some re-writing (good ol' pencil and paper tactics):
\begin{align}
&\sum_{k=0}^{m}\sum_{j=0}^{m-k} \frac{(-1)^j(x-a)^j}{k! j! (\frac{1}{m})^{j+k}}\\
&=\sum_{k=0}^{m}\sum_{j=0}^{m-k} \frac{(a-x)^jm^{j+k}}{k!j!}\\
&=\sum_{k=0}^{m}\sum_{j=k}^{m} \frac{(a-x)^{j-k}m^j}{k!(j-k)!}\\
&=\sum_{j=0}^{m}\sum_{k=0}^{j} \frac{(a-x)^{j-k}m^j}{k!(j-k)!}\\
&=\sum_{j=0}^{m}\frac{m^j}{j!}\sum_{k=0}^{j}\frac{j!}{k!(j-k)!}(a-x)^{j-k}1^k\\
&=\sum_{j=0}^{m}\frac{m^j}{j!}\sum_{k=0}^{j}{j\choose k}(a-x)^{j-k}1^k\\
&=\sum_{j=0}^{m}\frac{m^j}{j!}(a-x+1)^j\\
&=\sum_{j=0}^{m}\frac{(m(a-x+1))^j}{j!}\\
\end{align}
So that
$$\lim_{m\to\infty}\sum_{k=0}^{m}\sum_{j=0}^{m-k} \frac{(-1)^j(x-a)^j}{k! j! (\frac{1}{m})^{j+k}}=\lim_{m\to\infty}\sum_{j=0}^{m}\frac{(m(a-x+1))^j}{j!}$$
and knowing the taylor series of $e^x$, this is
$$\lim_{m\to\infty}\sum_{j=0}^{m}\frac{(m(a-x+1))^j}{j!}=\lim_{m\to\infty}e^{m(a-x+1)}$$
So your limit is divergent if $a-x+1>0$, it equals $1$ if $a-x+1=0$, and it equals $0$ if $a-x+1<0$.
