# Exponential Sequence is Uniformly Convergent on Compact Sets

## Theorem

Let $\EE = \sequence {E_n}$ denote the sequence of complex functions $E_n: \C \to \C$ defined as:

- $\map {E_n} z = \paren {1 + \dfrac z n}^n$

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

Then $\EE$ is uniformly convergent on $K$.

## Proof

### $\EE$ is Uniformly Bounded on an open space containing $K$

First, from Equivalence of Definitions of Complex Exponential Function we see that $\EE$ is pointwise convergent to $\exp$.

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From Combination Theorem for Continuous Complex 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:

- $M' = M + 1$
- $U = \set {z \in \C: \cmod z < M'}$
- $z \in U$
- $n \in \N$

From the Triangle Inequality for Complex Numbers:

- $\cmod {1 + \dfrac z n} \le 1 + \dfrac {\cmod z} n$

Thus, from Power Function is Strictly Increasing over Positive Reals: Natural Exponent:

- $\cmod {1 + \dfrac z n}^n \le \paren {1 + \dfrac {\cmod z} n}^n$

Call this result $(1)$.

Also:

\(\ds \cmod z\) | \(<\) | \(\ds M'\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds 1 + \frac {\cmod z} n\) | \(<\) | \(\ds 1 + \frac {M'} n\) | Divide both sides by $n$, add $1$ to both sides | ||||||||||

\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \paren {1 + \frac {\cmod z} n}^n\) | \(<\) | \(\ds \paren {1 + \frac {M'} n}^n\) | Power Function is Strictly Increasing over Positive Reals: Natural Exponent |

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

From Exponential Sequence is Eventually Increasing:

- $\exists N \in \N: n \ge N \implies \paren {1 + \dfrac {M'} n}^n \le \paren {1 + \dfrac {M'} {n + 1} }^{n + 1}$

So, for $n \ge N$:

\(\ds \paren {1 + \frac {M'} n}^n\) | \(\le\) | \(\ds \lim_{n \mathop \to \infty} \paren {1 + \frac {M'} n}^n\) | Limit of Bounded Convergent Sequence is Bounded | |||||||||||

\(\ds \) | \(=\) | \(\ds \exp M'\) |

Now, for each $n \in \set {1, 2, \ldots, N - 1}$:

- $\exists M_n \in \R: \paren {1 + \dfrac {M'} n}^n \le M_n$

by Continuous Function on Compact Subspace of Euclidean Space is Bounded.

So:

- $\forall n \in \N: \paren {1 + \dfrac {M'} n}^n \le \max \set {M_1, M_2, \ldots, M_{N - 1}, \exp M'}$

Call this result $(3)$.

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

\(\ds \forall n \in \N: \, \) | \(\ds \cmod {\paren {1 + \frac z n}^n}\) | \(=\) | \(\ds \cmod {1 + \dfrac z n}^n\) | Complex Modulus of Product of Complex Numbers | ||||||||||

\(\ds \) | \(\le\) | \(\ds \paren {1 + \dfrac {\cmod z} n}^n\) | From $(1)$ | |||||||||||

\(\ds \) | \(\le\) | \(\ds \paren {1 + \frac {M'} n}^n\) | From $(2)$ | |||||||||||

\(\ds \) | \(\le\) | \(\ds \max \set {M_1, M_2, \ldots, M_{N - 1}, \exp M'}\) | From $(3)$ |

That is, $\EE$ is uniformly bounded on $U$.

### $\EE$ is Uniformly Convergent on $K$

Note that, from Combination Theorem for Complex Derivatives:

- $\forall n \in \N: \map {E_n} z$ is holomorphic on $U$.

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

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

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

In particular, $\EE$ converges uniformly on $K$.

Hence the result.

$\blacksquare$