# 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:

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

$\blacksquare$