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.

$\blacksquare$