Null Sequence in Exponential Sequence/Proof 1

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

\(\ds \paren {1 + \frac {a_n} n}^n\)

\(=\)

\(\ds \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac {a_n} n}^k\)

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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$\blacksquare$