Lower Bound of Natural Logarithm/Proof 3
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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
Let $\sequence {f_n}$ be the sequence of mappings $f_n: \R_{>0} \to \R$ defined as:
- $\map {f_n} x = n \paren {\sqrt [n] x - 1 }$
Let $x \in \R_{>0}$ be fixed.
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: $0 < x < 1$
| \(\ds 0 < x < 1\) | \(\leadsto\) | \(\ds \forall k < n: \sqrt [n] x^{n - k} > x > 0\) | Power Function on Base between Zero and One is Strictly Decreasing/Rational Number | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds 0 < n x < \sum_{k = 0}^{n - 1} \sqrt [n] x^{n - k}\) | Real Number Ordering is Compatible with Addition | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \dfrac 1 {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} } > \dfrac 1 {n x}\) | Ordering of Reciprocals | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \dfrac {x - 1} {n x} < \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }\) | Order of Real Numbers is Dual of Order of their Negatives | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \dfrac {x - 1} {n x} < \sqrt [n] x - 1\) | Sum of Geometric Sequence | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds 1 - \dfrac 1 x < n \paren {\sqrt [n] x - 1}\) | Real Number Ordering is Compatible with Multiplication |
$\Box$
Case 2: $x = 1$
| \(\ds \dfrac {x - 1} x\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt [n] 1 - 1\) |
$\Box$
Case 3: $x > 1$
| \(\ds x > 1\) | \(\leadsto\) | \(\ds \forall k < n: 1 < \sqrt [n] x^{n - k} < x\) | Power Function on Base Greater than One is Strictly Increasing/Rational Number | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds 0 < \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k} < n x\) | Real Number Ordering is Compatible with Addition | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds 0 < \dfrac 1 {n x} < \dfrac 1 {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }\) | Ordering of Reciprocals | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \dfrac {x - 1} {n x} < \dfrac {x - 1} {1 + \sqrt [n] x + \sqrt [n] x^2 + \cdots + \sqrt [n] x^{n - 1} }\) | Real Number Ordering is Compatible with Multiplication | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds \dfrac {x - 1} {n x} < \sqrt [n] x - 1\) | Sum of Geometric Sequence | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds 1 - \dfrac 1 x < n \paren {\sqrt [n] x - 1 }\) | 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:
- $\ds 1 - \dfrac 1 x \le \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1 }$
from Limit of Bounded Convergent Sequence is Bounded.
Hence the result, from the definition of $\ln$.
$\blacksquare$
