Null Sequence in Exponential Sequence

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

Then:
 * $\displaystyle \lim_{n \to +\infty} \left(1 + \frac {a_n} n\right)^n = 1$

Proof 2
Let $\left \langle{E_n}\right \rangle$ be the sequence of functions $E_n: \C \to \C$ defined by $E_n(z) = \left({1 + \dfrac{z}{n}}\right)^n$.

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

Also note that $E_n \left({a_n}\right) = \left({1 + \dfrac{a_n}{n}}\right)^n$.

By Convergent Sequence in Metric Space is Bounded, we have that $\left \langle{a_n}\right \rangle$ is  bounded. Let this bound be $M$.

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

By Closed Disks are Compact, $K$ is  compact.

Since $K$ is compact, $\left \langle{E_n}\right \rangle$ is uniformly convergent on $K$.

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


 * $ \displaystyle \lim_{n \to \infty}E_n \left({a_n}\right) = \exp(0) = 1$

Hence the result.