Searching simpler proof for convergence of a sequence It is known that if $f_n \to f$ uniformly and $x_n \to x$ then $f_n(x_n) \to f(x)$. As an example, this can be applied in order to show that
$$\sum_{k=0}^n \frac{\left( 1-\frac{1}{n} \right)^k}{k!} \to e.$$
Now consider the sequence
$$z_n := \sum_{k=0}^n \frac{\left(1-\frac{k}{n}\right)^k}{k!}.$$
Since $z_n$ is dominated by $\sum_{k=0}^n \frac{1}{k!}$ it follows that $z_n$ converges to some $0 \leq c \leq e$.
I have shown by direct algebraic manipulations that $z_n \to e$.
Can the above theorem be also applied here in some way?
There is a double sequence $x_{n,k} := 1 - \frac{k}{n}$ which converges to $1$ for each fixed $k$. When setting $g_k(x) := \frac{x^k}{k!}$ then $z_n = \sum_{k=0}^n g_k(x_{n,k})$.
More generally, can the above theorem be appropriately generalized (e.g. $f$ is analytic and $f_n$ the first $n$ terms of the power series of $f$)?
 A: Solution
Given $ε \in (0,1)$:
$\def\nn{\mathbb{N}}$
  As $n \to \infty$:
    Let $m = \lfloor εn \rfloor$.
    Given $k \in \nn_{\le m}$:
      $(1-ε)^k \le (1-\frac{k}{n})^k$.
    Therefore $\displaystyle e(1-ε) - ε \le e^{1-ε} - \frac{1}{m!} \le \sum_{k=0}^m \frac{(1-ε)^k}{k!} \le \sum_{k=0}^n \frac{(1-\frac{k}{n})^k}{k!} \le \sum_{k=0}^n \frac{1}{k!} = e$.
      [because $1-ε \le e^{-ε}$ and $-ε \le \dfrac{1}{m!}$ since $n \to \infty$.]
  Therefore $\displaystyle e - (e+1)ε \le \liminf_{n\to\infty} \sum_{k=0}^n \frac{(1-\frac{k}{n})^k}{k!} \le \limsup_{n\to\infty} \sum_{k=0}^n \frac{(1-\frac{k}{n})^k}{k!} \le e$
Therefore $\displaystyle \lim_{n\to\infty} \sum_{k=0}^n \frac{(1-\frac{k}{n})^k}{k!} = e$.
Notes
Notice that I used limit inferior/superior above because we do not know the existence of the limit. This technique is extremely elementary but very powerful and can be used in almost all such limits, including those that involve limits of integrals. Other people often use things like the Dominated Convergence Theorem, but unless in completely abstract settings that is an unnecessarily heavy hammer. Also, in some cases it is troublesome to have to manipulate into a form on which the theorems work.
