Null Sequence in Exponential Sequence

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

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


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

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


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.