Null Sequence in Exponential Sequence
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 1
\(\ds \paren {1 + \frac {a_n} n}^n\) | \(=\) | \(\ds \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac {a_n} n}^k\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds {n \choose 0} \paren {\frac {a_n} n}^0 + \sum_{k \mathop = 1}^n {n \choose k} \paren {\frac {a_n} n}^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + a_n \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{n \mathop \to +\infty} 1 + a_n \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}\) | \(=\) | \(\ds 1 + \paren {\lim_{n \mathop \to +\infty} a_n} \paren {\lim_{n \mathop \to +\infty} \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k} }\) | Combination Theorem for Sequences | ||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 0 \cdot \paren {\lim_{n \mathop \to +\infty} \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
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$\blacksquare$
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:
- $\ds \lim_{n \mathop \to \infty} \map {E_n} z = \map \exp z$
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.
$\blacksquare$