Nth Derivative of Natural Logarithm by Reciprocal

Theorem

 * $\dfrac {\rd^n} {\rd x^n} \dfrac {\ln x} x = \left({-1}\right)^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$

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

Proof
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\dfrac {\rd^n} {\rd x^n} \dfrac {\ln x} x = \left({-1}\right)^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$

Basis for the Induction
$P \left({1}\right)$ is the case:
 * $\dfrac \rd {\rd x} \dfrac {\ln x} x = \left({-1}\right)^n n! \dfrac {H_n - \ln x} {x^{n + 1} }$

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\dfrac {\rd^k} {\rd x^k} \dfrac {\ln x} x = \left({-1}\right)^{k + 1} k! \dfrac {H_k - \ln x} {x^{k + 1} }$

from which it is to be shown that:
 * $\dfrac {\rd^{k + 1} } {\rd x^{k + 1} } \dfrac {\ln x} x = \left({-1}\right)^{k + 2} \left({k + 1}\right)! \dfrac {H_{k + 1} - \ln x} {x^{k + 2} }$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{>0}: \dfrac {\rd^n} {\rd x^n} \dfrac {\ln x} x = \left({-1}\right)^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$