# Lower Bound of Natural Logarithm

## Theorem

$\forall x \in \R_{>0}: 1 - \dfrac 1 x \le \ln x$

where $\ln x$ denotes the natural logarithm of $x$.

## Proof 1

Let $x > 0$.

 $\displaystyle x - 1$ $\ge$ $\displaystyle \ln x$ Upper Bound of Natural Logarithm $\displaystyle \implies \ \$ $\displaystyle \frac 1 x -1$ $\ge$ $\displaystyle \ln \frac 1 x$ putting $\frac 1 x$ into the above inequality $\displaystyle$ $=$ $\displaystyle - \ln x$ Logarithm of Reciprocal $\displaystyle \implies \ \$ $\displaystyle 1 - \frac 1 x$ $\le$ $\displaystyle \ln x$ multiplying throughout by $-1$

$\blacksquare$

## Proof 2

Let $x > 0$.

Note that:

$1 - \dfrac 1 x \le \ln x$

is logically equivalent to:

$1 - \dfrac 1 x - \ln x \le 0$

Let $f \left({x}\right) = 1 - \dfrac 1 x - \ln x$.

Then:

 $\displaystyle f \left({x}\right)$ $=$ $\displaystyle 1 - \dfrac 1 x - \ln x$ $\displaystyle \implies \ \$ $\displaystyle f' \left({x}\right)$ $=$ $\displaystyle \frac 1 {x^2} - \frac 1 x$ Derivative of Constant, Power Rule for Derivatives, Derivative of Natural Logarithm Function $\displaystyle$ $=$ $\displaystyle \frac {1 - x} {x^2}$ $\displaystyle \implies \ \$ $\displaystyle f'' \left({x}\right)$ $=$ $\displaystyle - \frac 2 {x^3} + \frac 1 {x^2}$ Power Rule for Derivatives

Note that $f' \left({1}\right) = 0$.

Also, $f'' \left({1}\right) < 0$.

So by the Second Derivative Test, $x = 1$ is a local maximum.

On $\left ({0 \,.\,.\, 1} \right)$:

$f' \left({x}\right) > 0$

By Derivative of Monotone Function, $f$ is strictly increasing on that interval.

On $\left ({1 \,.\,.\, +\infty} \right)$:

$f'\left({x}\right) < 0$

By Derivative of Monotone Function, $f$ is strictly decreasing on that interval.

So $x = 1$ yields a global maximum, at which by Logarithm of 1 is 0:

$f \left({1}\right) = 1 - 1 - 0 = 0$

That is:

$\forall x > 0: f \left({x}\right) \le 0$

and so by definition of $f \left({x}\right)$:

$1 - \dfrac 1 x - \ln x \le 0$

$\blacksquare$

## Proof 3

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 $n \in \N$.

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

### 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:

$\forall n \in \N : 1 - \dfrac{1}{x} \leq n\left({ \sqrt[n]x - 1 }\right)$

Thus:

$\displaystyle 1 - \dfrac{1}{x} \leq \lim_{n \to \infty} n\left({ \sqrt[n]x - 1 }\right)$

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

$\blacksquare$

## Illustration 