Lower Bound of Natural Logarithm/Proof 3

Theorem

$\forall x \in \R_{>0}: 1 - \dfrac 1 x \le \ln x$

where $\ln x$ denotes the natural logarithm of $x$.

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 $n \in \N$.

$\sqrt [n] x - 1 = \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }$

Case 1: $0 < x < 1$

 $\displaystyle 0 < x < 1$ $\leadsto$ $\displaystyle \forall k < n: \sqrt [n] x^{n - k} > x > 0$ Power Function on Base between Zero and One is Strictly Decreasing/Rational Number $\displaystyle$ $\leadsto$ $\displaystyle 0 < n x < \sum_{k = 0}^{n - 1} \sqrt [n] x^{n - k}$ Real Number Ordering is Compatible with Addition $\displaystyle$ $\leadsto$ $\displaystyle \dfrac 1 {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} } > \dfrac 1 {n x}$ Ordering of Reciprocals $\displaystyle$ $\leadsto$ $\displaystyle \dfrac {x - 1} {n x} < \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }$ Order of Real Numbers is Dual of Order of their Negatives $\displaystyle$ $\leadsto$ $\displaystyle \dfrac {x - 1} {n x} < \sqrt [n] x - 1$ Sum of Geometric Sequence $\displaystyle$ $\leadsto$ $\displaystyle 1 - \dfrac 1 x < n \paren {\sqrt [n] x - 1}$ Real Number Ordering is Compatible with Multiplication

$\Box$

Case 2: $x = 1$

 $\displaystyle \dfrac {x - 1} x$ $=$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle \sqrt [n] 1 - 1$

$\Box$

Case 3: $x > 1$

 $\displaystyle x > 1$ $\leadsto$ $\displaystyle \forall k < n: 1 < \sqrt [n] x^{n - k} < x$ Power Function on Base Greater than One is Strictly Increasing/Rational Number $\displaystyle$ $\leadsto$ $\displaystyle 0 < \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k} < n x$ Real Number Ordering is Compatible with Addition $\displaystyle$ $\leadsto$ $\displaystyle 0 < \dfrac 1 {n x} < \dfrac 1 {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }$ Ordering of Reciprocals $\displaystyle$ $\leadsto$ $\displaystyle \dfrac {x - 1} {n x} < \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }$ Real Number Ordering is Compatible with Multiplication $\displaystyle$ $\leadsto$ $\displaystyle \dfrac {x - 1} {n x} < \sqrt [n] x - 1$ Sum of Geometric Sequence $\displaystyle$ $\leadsto$ $\displaystyle 1 - \dfrac 1 x < n \paren {\sqrt [n] x - 1 }$ Real Number Ordering is Compatible with Multiplication

$\Box$

Thus:

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

Thus:

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

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

$\blacksquare$