# 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 $\left\langle{ f_n }\right\rangle$ 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 $n \in \N$.

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

### Case 1: $0 < x < 1$

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

$\Box$

### Case 2: $x = 1$

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

$\Box$

### Case 3: $x > 1$

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

$\Box$

Thus:

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

Thus:

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

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

$\blacksquare$