User:Keith.U/Sandbox

Theorem
Let $\mathcal E = \left \langle{E_n}\right \rangle$ denote the sequence of functions $E_n: \C \to \C$ defined as:
 * $E_n \left({z}\right) = \left({1 + \dfrac z n}\right)^n$

Let $K$ be a compact subset of $\C$.

Then $\mathcal E$ is uniformly convergent on $K$.

$\mathcal E$ is Uniformly Bounded on an open space containing $K$
First, from Equivalence of Definitions of Complex Exponential we see that $\mathcal E$ is pointwise convergent to $\exp$.

From Combination Theorem for Continuous Functions, $E_n$ is continuous for each $n \in \N$.

From Compact Subspace of Metric Space is Bounded, $K$ is bounded by some real number $M \in \R$.

Let:
 * $M' = M + 1$
 * $U = \left \{{z \in \C: \left \vert{z}\right \vert < M'}\right \}$
 * $z \in U$
 * $n \in \N$

From the Triangle Inequality for Complex Numbers:
 * $\left \vert{1 + \dfrac z n}\right \vert \le 1 + \dfrac {\left \vert{z}\right \vert} n$

Thus, from Power Function is Strictly Increasing over Positive Reals: Natural Exponent:
 * $\left \vert{1 + \dfrac z n}\right \vert^n \le \left({1 + \dfrac {\left \vert{z}\right \vert} n}\right)^n$

Call this result $(1)$.

Also:

Thus it is sufficient to show that $\left({1 + \dfrac {M'} n}\right)^n$ is bounded.

From Exponential Sequence is Eventually Increasing:
 * $\exists N \in \N: n \geq N \implies \left({1 + \dfrac {M'} n}\right)^n \le \left({1 + \dfrac {M'} {n + 1}}\right)^{n + 1}$

So, for $n \geq N$:

Now, for each $n \in \left \{{1, 2, \ldots, N-1}\right \}$:
 * $\exists M_n \in \R: \left({1 + \dfrac {M'} n}\right)^n \leq M_n$

by Continuous Function on Compact Space is Bounded.

So:
 * $\forall n \in \N: \left({1 + \dfrac {M'} n}\right)^n \le \max \left \{{M_1, M_2, \ldots, M_{N-1}, \exp M'}\right \}$

Call this result $(3)$.

Hence, for $z \in U$; that is, for $\left \vert{z}\right \vert < {M'}$:

That is, $\mathcal E$ is uniformly bounded on $U$.

$\mathcal E$ is Uniformly Convergent on $K$
Note that, from Combination Theorem for Complex Derivatives:
 * $\forall n \in \N: E_n \left({z}\right)$ is holomorphic on $U$.

From Uniformly Bounded Family is Locally Bounded, $\mathcal E$ is locally bounded on $U$.

From Montel's Theorem, $\mathcal E$ is a normal family.

From Vitali's Convergence Theorem, $\mathcal E$ converges uniformly on any compact subset of $U$.

In particular, $\mathcal E$ converges uniformly on $K$.

Hence the result.