Null Sequence in Exponential Sequence/Proof 1
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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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$\blacksquare$