# Lower Bound of Natural Logarithm/Proof 3

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## Contents

## 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 $n \in \N$.

From Sum of Geometric Sequence:

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

### 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:

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

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 $\ln$.

$\blacksquare$