Forward Difference of Harmonic Number Function
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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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^x \frac 1 k + \frac 1 {x + 1} - \sum_{k \mathop = 1}^x \frac 1 k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {x + 1}\) |
$\blacksquare$