Null Sequence in Exponential Sequence

Theorem
Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a sequence of complex numbers such that:
 * $\displaystyle \lim_{n \mathop \to +\infty}a_n = 0$

Then:
 * $\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$

Proof 2
Let $\sequence {E_n}$ be the sequence of functions $E_n: \C \to \C$ defined by $E_n \paren z = \paren {1 + \dfrac z n}^n$.

Note that $\displaystyle \lim_{n \mathop \to \infty} E_n \paren z = \exp \paren z$, where $\exp \paren z$ is the complex exponential.

Also note 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:


 * $ \displaystyle \lim_{n \mathop \to \infty} E_n \paren {a_n} = \exp \paren 0 = 1$

Hence the result.