Derivative of Complex Power Series/Proof 2/Lemma

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

Then:
 * $\displaystyle \lim_{n \to \infty} \left[{ \dfrac {n \left({n-1}\right)}{2} }\right]^{1/n} = 1$

Proof
Choose any $\alpha > 1$.

It follows from the ratio test that:


 * $\displaystyle \lim_{n \to \infty} \dfrac{1}{\alpha^n} \frac{n \left({n-1}\right) }{2}=0$

Therefore, for all sufficiently large $n$,


 * $\displaystyle \dfrac {n \left({n-1}\right) }{2} \le \alpha^n$

and so:


 * $\displaystyle \left[{ \dfrac {n \left({ n-1 }\right) }{2} }\right]^{1/n} \le \alpha$

It follows that:


 * $\displaystyle \lim_{n \to \infty} \left[{ \dfrac {n \left({n-1}\right) }{2} }\right]^{1/n} \le \alpha$

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


 * $\displaystyle \lim_{n \to \infty} \left[{ \dfrac {n \left({n-1}\right) }{2} }\right]^{1/n} \le 1$

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


 * $\displaystyle \left[{ \dfrac {n \left({n-1}\right) }{2} }\right]^{1/n} \ge 1^{1/n} = 1$

Therefore:


 * $\displaystyle \lim_{n \to \infty} \left[{ \dfrac {n \left({n-1}\right) }{2} }\right]^{1/n} = 1$