Derivative of Complex Power Series/Proof 2/Lemma

Lemma
Let $n \in N_{\ge 1}$.

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$

Proof
Choose any $\alpha > 1$.

It follows from the ratio test that:


 * $\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 {\alpha^n} \frac {n \paren {n - 1} } 2 = 0$

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:


 * $\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le \alpha$

Since $\alpha > 1$ was arbitrary, we can conclude that:


 * $\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \le 1$

It is clear that the following holds for sufficiently large $n$:


 * $\displaystyle \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} \ge 1^{1/n} = 1$

Therefore:


 * $\displaystyle \lim_{n \mathop \to \infty} \paren {\dfrac {n \paren {n - 1} } 2}^{1/n} = 1$