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

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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

$\blacksquare$


Sources

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