Show that $\lim_{n\rightarrow\infty}(\sum_{m=1}^{\infty}S_{m,n})=\sum_{m=1}^{\infty}(\lim_{n\rightarrow\infty}S_{m,n}).$ Let $m,n\in \mathbf{Z^{+}}$,
$$S_{m,n}=\begin{cases}
 (1-\frac{m}{n})^{n} ,\quad\text{if} \quad  1 \leq m < n\\ 
 \qquad 0,\qquad\text{if}\quad m\geq n
\end{cases}.$$
Show that $\lim_{n\rightarrow\infty}(\sum_{m=1}^{\infty}S_{m,n})=\sum_{m=1}^{\infty}(\lim_{n\rightarrow\infty}S_{m,n}).$
We know that $\lim_{n\rightarrow\infty}S_{m,n}=e^{-m},$ So $\sum_{m=1}^{\infty}(\lim_{n\rightarrow\infty}S_{m,n})=\sum_{m=1}^{\infty}e^{-m}=\frac{1}{e-1}.$ But why the signs $\lim$ and $\sum$ can interchange?
 A: Using Bernoulli's inequality with $m < n$ and $-m/n > -1$,
$$\left(1 - \frac{m}{n+1}\right)^{\frac{n+1}{n}} \geqslant 1 - \frac{n+1}{n}\frac{m}{n+1} = 1 - \frac{m}{n} \\ \implies\left(1 - \frac{m}{n+1}\right)^{n+1} \geqslant \left(1 - \frac{m}{n}\right)^{n}.$$
Thus  $S_{m,n} \uparrow e^{-m}$ as $n \to \infty.$
Now apply the monotone convergence theorem for series.
A: By The Sum Law for Limits:
$$ \text{If }\space \lim_{x\rightarrow\alpha}f(x) \space\text{ and }\space \lim_{x\rightarrow\alpha}g(x) \space\text{ both exist } \space\Rightarrow\space \lim_{x\rightarrow\alpha}\left(f(x)+g(x)\right) = \lim_{x\rightarrow\alpha}f(x) + \lim_{x\rightarrow\alpha}g(x) $$
And because $\{S_{m,n}=0\colon{m}\ge{n}\}\space\Rightarrow$ 
$$
\lim_{n\rightarrow\alpha}\,\sum_{m=0}^{\infty}S_{m,n}=\lim_{n\rightarrow\alpha}\left(\sum_{m=0}^{\alpha}S_{m,n}+\sum_{m=\alpha+1}^{\infty}S_{m,n}\right) = \\[6mm]
\color{red}{\lim_{n\rightarrow\alpha}\,\sum_{m=0}^{\alpha}S_{m,n}} = \lim_{n\rightarrow\alpha}\,\sum_{m=0}^{\alpha}\left(1-\frac{m}{n}\right)^{n} = \small \lim_{n\rightarrow\alpha}\left[\left(1-\frac{0}{n}\right)^{n}+\left(1-\frac{1}{n}\right)^{n}+\cdots+\left(1-\frac{\alpha}{n}\right)^{n}\right] \normalsize = \\[6mm]
\small \lim_{n\rightarrow\alpha}\left(1-\frac{0}{n}\right)^{n}+\lim_{n\rightarrow\alpha}\left(1-\frac{1}{n}\right)^{n}+\cdots+\lim_{n\rightarrow\alpha}\left(1-\frac{\alpha}{n}\right)^{n} \normalsize = \sum_{m=0}^{\alpha}\,\lim_{n\rightarrow\alpha}\left(1-\frac{m}{n}\right)^{n} = \color{red}{\sum_{m=0}^{\alpha}\,\lim_{n\rightarrow\alpha}S_{m,n}}
$$
Thus, the actual starting point is through ($\alpha$), not immediately ($\infty$)! 
And now, Let $(\alpha\rightarrow\infty)\space\Rightarrow$ 
$$ \lim_{\alpha\rightarrow\infty}\left[\lim_{n\rightarrow\alpha}\,\sum_{m=0}^{\alpha}S_{m,n}\right] = \lim_{\alpha\rightarrow\infty}\left[\sum_{m=0}^{\alpha}\,\lim_{n\rightarrow\alpha}S_{m,n}\right] \quad\Rightarrow\quad \color{red}{\lim_{n\rightarrow\infty}\,\sum_{m=0}^{\infty}S_{m,n} = \sum_{m=0}^{\infty}\,\lim_{n\rightarrow\infty}S_{m,n}}$$
In Other words, the trick is to use an intermediate variable, which should allowed you to start from the inside to the outside and then take ONE limit as a final stage at the END. You DO NOT take the inner limit then the outer limit (Thanks Simple Art).
