# Summation to n of Reciprocal of k by k-1 of Harmonic Number

## Theorem

$\ds \sum_{1 \mathop < k \mathop \le n} \dfrac 1 {k \paren {k - 1} } H_k = 2 - \dfrac {H_n} n - \dfrac 1 n$

where $H_n$ denotes the $n$th harmonic number.

## Proof

 $\ds \sum_{1 \mathop < k \mathop \le n} \dfrac 1 {k \paren {k - 1} } H_k$ $=$ $\ds \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {\paren {k + 1} k} H_{k + 1}$ Translation of Index Variable of Summation $\ds$ $=$ $\ds -\sum_{k \mathop = 1}^{n - 1} \paren {\dfrac 1 {k + 1} - \dfrac 1 k} H_{k + 1}$ $\ds$ $=$ $\ds -\paren {\dfrac {H_{n + 1} } n - \dfrac {H_2} 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1} \paren {H_{k + 2} - H_{k + 1} } }$ Abel's Lemma: Formulation 1 $\ds$ $=$ $\ds H_2 - \dfrac {H_{n + 1} } n + \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1} \dfrac 1 {k + 2}$ simplification $\ds$ $=$ $\ds \frac 3 2 - \dfrac {H_{n + 1} } n + \sum_{k \mathop = 1}^{n - 1} \paren {\dfrac 1 {k + 1} - \dfrac 1 {k + 2} }$ separating the fractions, and Harmonic Number $H_2$ $\ds$ $=$ $\ds \frac 3 2 - \dfrac {H_{n + 1} } n + \frac 1 2 - \dfrac 1 {n + 1}$ Telescoping Series $\ds$ $=$ $\ds 2 - \frac 1 n \paren {H_n + \dfrac 1 {n + 1} } - \dfrac 1 {n + 1}$ $\ds$ $=$ $\ds 2 - \frac {H_n} n - \dfrac 1 {n \paren {n + 1} } - \dfrac 1 {n + 1}$ $\ds$ $=$ $\ds 2 - \frac {H_n} n - \paren {\dfrac 1 n - \dfrac 1 {n + 1} } - \dfrac 1 {n + 1}$ $\ds$ $=$ $\ds 2 - \frac {H_n} n - \dfrac 1 n$ after simplification

$\blacksquare$