Derivative of Complex Power Series/Proof 2/Lemma
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Lemma
Let $n \in N_{\ge 1}$.
Then:
- $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$
Proof
Choose any $\alpha > 1$.
It follows from the ratio test that:
- $\ds \lim_{n \mathop \to \infty} \dfrac 1 {\alpha^n} \frac {n \paren {n - 1} } 2 = 0$
Therefore, for all sufficiently large $n$:
- $\dfrac {n \paren {n - 1} } 2 \le \alpha^n$
and so:
- $\paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le \alpha$
It follows that:
- $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le \alpha$
Since $\alpha > 1$ was arbitrary, we can conclude that:
- $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le 1$
It is clear that the following holds for sufficiently large $n$:
- $\ds \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \ge 1^{1/n} = 1$
Therefore:
- $\ds \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$
$\blacksquare$