Null Sequence in Exponential Sequence/Proof 2

Theorem

Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a sequence of complex numbers such that:

$\ds \lim_{n \mathop \to +\infty}a_n = 0$

Then:

$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$

Proof

Let $\sequence {E_n}$ be the sequence of complex functions $E_n: \C \to \C$ defined by:

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

We have that:

$\ds \lim_{n \mathop \to \infty} \map {E_n} z = \map \exp z$

where $\map \exp z$ is the complex exponential.

We also have that:

$E_n \paren {a_n} = \paren {1 + \dfrac {a_n} n}^n$

By Convergent Sequence in Metric Space is Bounded, we have that $\sequence {a_n}$ is Bounded Complex Sequence.

Let this bound be $M$.

Let $K \subseteq \C$ be the closed disk of radius $M$.

By Closed Disk is Compact, $K$ is compact.

By Exponential Sequence is Uniformly Convergent on Compact Sets, $\sequence {E_n}$ is uniformly convergent on $K$.

Now the hypotheses of Uniformly Convergent Sequence Evaluated on Convergent Sequence are satisfied, so:

$\ds \lim_{n \mathop \to \infty} \map {E_n} {a_n} = \map \exp 0 = 1$

Hence the result.

$\blacksquare$