Defining Sequence of Natural Logarithm is Strictly Decreasing
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Theorem
Let $x \in \R$ be a real number such that $x > 0$.
Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:
- $\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Then $\forall x \in \R_{>0}: \sequence {\map {f_n} x}$ is strictly decreasing.
Proof
Fix $t \in R_{>0}$.
Then:
\(\ds n \paren {t^{n + 1} - 1} - \paren {n + 1} \paren {t^n - 1}\) | \(=\) | \(\ds \paren {t - 1}^2 \paren {1 + 2 t + 3 t^2 + \ldots + n t^{n - 1} }\) | Sum of Geometric Sequence: Corollary 2 | |||||||||||
\(\ds \) | \(>\) | \(\ds 0\) | Product of Real Numbers is Positive iff Numbers have Same Sign |
Thus:
- $n \paren {t^{n + 1} - 1} > \paren {n + 1} \paren {t^n - 1}$
Fix $x \in \R_{>0}$.
From Power of Positive Real Number is Positive: Rational Number:
- $\forall n \in \N : x^{1 / \paren {n \paren {n + 1} } } \in \R_{>0}$
Thus:
\(\ds \forall t \in \R_{>0}: \, \) | \(\ds n \paren {t^{n + 1} - 1}\) | \(>\) | \(\ds \paren {n + 1} \paren {t^n - 1}\) | by the result obtained above | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds n \paren {\paren {x^{1 / \paren {n \paren {n + 1} } } }^{n + 1} - 1}\) | \(>\) | \(\ds \paren {n + 1} \paren {\paren {x^{1 / \paren {n \paren {n + 1} } } }^n - 1 }\) | as $x^{1 / \paren {n \paren {n + 1} } } \in \R_{>0}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds n \paren {x^{1 / n} - 1}\) | \(>\) | \(\ds \paren {n + 1} \paren {x^{1 / \paren {n + 1} } - 1}\) | Product of Indices of Real Number: Rational Numbers | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds n \paren {\sqrt [n] x - 1}\) | \(>\) | \(\ds \paren {n + 1} \paren {\map {f_n} x - 1 }\) | Definition of $n$th root |
Hence the result, by definition of strictly decreasing sequence.
$\blacksquare$