Lower Bound of Natural Logarithm/Proof 3

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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 $n \in \N$.

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 1: $0 < x < 1$

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

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

$\Box$

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 $\ln$.

$\blacksquare$


Illustration

LowerBoundForLn.png