# Upper Bound of Natural Logarithm

## 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

$\ln$ is (strictly) concave.
$\ln y - \ln 1 \le \left({D \ln 1}\right) \left({y - 1}\right)$
$D \ln 1 = \dfrac 1 1 = 1$

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:

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

Thus:

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

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

$\blacksquare$