Lower Bound of Natural Logarithm/Proof 3

Proof
Let $\left\langle{ f_n }\right\rangle$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:
 * $f_n \left({ x }\right) = n \left({ \sqrt[n]{ x } - 1 }\right)$

Fix $x \in \R_{>0}$.

We first show that $\forall n \in \N : 1 - \dfrac{1}{x} \leq < n\left({ \sqrt[n]x - 1 }\right)$

Let $n \in \N$.

From Sum of Geometric Progression:
 * $ \sqrt[n]x - 1 = \dfrac{ x - 1 }{ 1 + \sqrt[n]x + \sqrt[n]x^{2} + \cdots + \sqrt[n]x^{n - 1} }$

Case 3: $x > 1$
Thus:
 * $\forall n \in \N : 1 - \dfrac{1}{x} \leq n\left({ \sqrt[n]x - 1 }\right)$

by Proof by Cases.

Thus:
 * $ \displaystyle 1 - \dfrac{1}{x} \leq \lim_{n \to \infty} n\left({ \sqrt[n]x - 1 }\right) $

from Limit of Bounded Convergent Sequence is Bounded.

Hence the result, from the definition of $\ln$.