# 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 1

 $\displaystyle \paren {1 + \frac {a_n} n}^n$ $=$ $\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac {a_n} n}^k$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle {n \choose 0} \paren {\frac {a_n} n}^0 + \sum_{k \mathop = 1}^n {n \choose k} \paren {\frac {a_n} n}^k$ $\displaystyle$ $=$ $\displaystyle 1 + a_n \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}$ $\displaystyle \leadsto \ \$ $\displaystyle \displaystyle \lim_{n \mathop \to +\infty} 1 + a_n \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}$ $=$ $\displaystyle 1 + \paren \lim_{n \mathop \to +\infty} a_n} \paren \lim_{n \mathop \to +\infty} \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}$ Combination Theorem for Sequences $\displaystyle$ $=$ $\displaystyle 1 + 0 \cdot \paren \lim_{n \mathop \to +\infty} \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}$ $\displaystyle$ $=$ $\displaystyle 1$

$\blacksquare$

## Proof 2

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:

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

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:

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

Hence the result.

$\blacksquare$