Forward Difference of Harmonic Number Function

Theorem

Let $H_x$ denote the harmonic number function.

Then:

$\Delta H_x = \dfrac 1 {x + 1}$

where $\Delta H_x$ denotes the forward difference operator.

Proof

From the definitions:

\(\ds \Delta H_x\)

\(=\)

\(\ds H_{x + 1} - H_x\)

Definition of Forward Difference Operator

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 1}^{x + 1} \frac 1 k - \sum_{k \mathop = 1}^x \frac 1 k\)

Definition of Harmonic Number

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 1}^x \frac 1 k + \frac 1 {x + 1} - \sum_{k \mathop = 1}^x \frac 1 k\)

\(\ds \)

\(=\)

\(\ds \frac 1 {x + 1}\)

$\blacksquare$