Lower Bound of Natural Logarithm

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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 1

Let $x > 0$.

\(\displaystyle x - 1\) \(\ge\) \(\displaystyle \ln x\) Upper Bound of Natural Logarithm
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac 1 x -1\) \(\ge\) \(\displaystyle \ln \frac 1 x\) putting $\frac 1 x$ into the above inequality
\(\displaystyle \) \(=\) \(\displaystyle -\ln x\) Logarithm of Reciprocal
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1 - \frac 1 x\) \(\le\) \(\displaystyle \ln x\) 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 $\sequence {f_n}$ be 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 $\forall n \in \N : 1 - \dfrac 1 x \le < n \paren {\sqrt [n] x - 1}$

Let $n \in \N$.

From Sum of Geometric Sequence:

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


Case 1: $0 < x < 1$

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

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

$\Box$


Thus:

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

by Proof by Cases.

Thus:

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

from Limit of Bounded Convergent Sequence is Bounded.

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

$\blacksquare$


Illustration

LowerBoundForLn.png


Also see