Lower Bound of Natural Logarithm/Proof 3

Proof
Let $\sequence {f_n}$ be the sequence of mappings $f_n: \R_{>0} \to \R$ defined as:
 * $\map {f_n} x = n \paren {\sqrt [n] x - 1 }$

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

We first show that $\forall n \in \N : 1 - \dfrac 1 x \le < n \paren {\sqrt [n] x - 1}$

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:
 * $\forall n \in \N : 1 - \dfrac 1 x \le n \paren {\sqrt [n] x - 1 }$

by Proof by Cases.

Thus:
 * $\displaystyle 1 - \dfrac 1 x \le \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1 }$

from Limit of Bounded Convergent Sequence is Bounded.

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