Upper Bound of Natural Logarithm/Proof 2

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Theorem

Let $\ln x$ be the natural logarithm of $x$ where $x \in \R_{>0}$.


Then:

$\ln x \le x - 1$


Proof

Let $\sequence {f_n}$ denote 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 : n \paren {\sqrt [n] x - 1} < x - 1 $


Case 1: $0 < x < 1$

Suppose $0 < x < 1$.

Then:

\(\ds 0\) \(<\) \(\, \ds x \, \) \(\, \ds < \, \) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(<\) \(\, \ds \sqrt [n] x^{n - k} \, \) \(\, \ds < \, \) \(\ds 1\) Power Function on Base between Zero and One is Strictly Decreasing/Rational Number \(\quad\) $\forall k \in \set {0, 1, \ldots, n - 1}$
\(\ds \leadsto \ \ \) \(\ds 0\) \(<\) \(\, \ds \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k} \, \) \(\, \ds < \, \) \(\ds n\) Real Number Ordering is Compatible with Addition
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\, \ds \frac 1 n \, \) \(\, \ds < \, \) \(\ds \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) Ordering of Reciprocals
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\, \ds 1 \, \) \(\, \ds < \, \) \(\ds \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) multiplying both sides by $n$
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\, \ds x - 1 \, \) \(\, \ds > \, \) \(\ds \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) multiplying both sides by $x - 1 < 0$
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\, \ds y^n - 1 \, \) \(\, \ds > \, \) \(\ds \frac {\paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }\) substituting $y = \sqrt [n] x$
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\, \ds y^n - 1 \, \) \(\, \ds > \, \) \(\ds n \paren {y - 1}\) Sum of Geometric Sequence
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\, \ds x - 1 \, \) \(\, \ds > \, \) \(\ds n \paren {\sqrt [n] x - 1}\) substituting $\sqrt [n] x = y$

$\Box$


Case 2: $x = 1$

Suppose $x = 1$.

Then:

\(\ds \map \ln x\) \(=\) \(\ds \map \ln 1\)
\(\ds \) \(=\) \(\ds 0\) Natural Logarithm of 1 is 0/Proof 3
\(\ds \) \(=\) \(\ds 1 - 1\)
\(\ds \) \(=\) \(\ds x - 1\)

$\Box$


Case 3: $x > 1$

Suppose $x > 1$.

Then:

\(\ds x\) \(>\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \sqrt [n] x^{n - k}\) \(>\) \(\ds 1\) Power Function on Base Greater than One is Strictly Increasing/Rational Number \(\quad\) $\forall k \in \set { 0, 1, \ldots, n - 1}$
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k}\) \(>\) \(\ds n\) Real Number Ordering is Compatible with Addition
\(\ds \leadsto \ \ \) \(\ds \frac 1 n\) \(>\) \(\ds \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) Ordering of Reciprocals
\(\ds \leadsto \ \ \) \(\ds 1\) \(>\) \(\ds \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) Multiply both sides by $n$
\(\ds \leadsto \ \ \) \(\ds x - 1\) \(>\) \(\ds \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) multiplying both sides by $x - 1 > 1$
\(\ds \leadsto \ \ \) \(\ds y^n - 1\) \(>\) \(\ds \frac {n \paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }\) substituting $y = \sqrt [n] x$
\(\ds \leadsto \ \ \) \(\ds y^n - 1\) \(>\) \(\ds n \paren {y - 1}\) Sum of Geometric Sequence
\(\ds \leadsto \ \ \) \(\ds x - 1\) \(>\) \(\ds n \paren {\sqrt [n] x - 1}\) Substitute $\sqrt [n] x = y$

$\Box$


Thus:

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

by Proof by Cases.


Thus:

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

from Limit of Bounded Convergent Sequence is Bounded.


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

$\blacksquare$