Nth Derivative of Natural Logarithm by Reciprocal

Theorem

 * $\dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$

where $H_n$ denotes the $n$th harmonic number:
 * $H_n = \ds \sum_{r \mathop = 1}^n \dfrac 1 r = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 r$

Proof
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $\dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$

Basis for the Induction
$\map P 1$ is the case:
 * $\dfrac \d {\d x} \dfrac {\ln x} x = \paren {-1}^n n! \dfrac {H_n - \ln x} {x^{n + 1} }$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\dfrac {\d^k} {\d x^k} \dfrac {\ln x} x = \paren {-1}^{k + 1} k! \dfrac {H_k - \ln x} {x^{k + 1} }$

from which it is to be shown that:
 * $\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac {\ln x} x = \paren {-1}^{k + 2} \paren {k + 1}! \dfrac {H_{k + 1} - \ln x} {x^{k + 2} }$

Induction Step
This is the induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{>0}: \dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$