Upper Bound of Natural Logarithm

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Theorem

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


Then:

$\ln y \le y - 1$


Corollary

$\forall s \in \R_{>0}: \ln x \le \dfrac {x^s} s$


Proof 1

From Logarithm is Strictly Concave:

$\ln$ is (strictly) concave.

From Mean Value of Concave Real Function:

$\ln y - \ln 1 \le \left({D \ln 1}\right) \left({y - 1}\right)$

From Derivative of Natural Logarithm:

$D \ln 1 = \dfrac 1 1 = 1$

So:

$\ln y - \ln 1 \le \left({y - 1}\right)$

But from Logarithm of 1 is 0:

$\ln 1 = 0$

Hence the result.

$\blacksquare$


Proof 2

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:

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

$\Box$


Case 2: $x = 1$

Suppose $x = 1$.

Then:

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

$\Box$


Case 3: $x > 1$

Suppose $x > 1$.

Then:

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

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

from Limit of Bounded Convergent Sequence is Bounded.


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

$\blacksquare$


Also see