Summation to n of kth Harmonic Number over k

Theorem

$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} k = \dfrac { {H_n}^2 + \harm 2 n } 2$

where:

$H_n$ denotes the $n$th harmonic number

$\harm 2 n$ denotes the general harmonic number of order $2$ evaluated at $n$.

$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} k$

$=$

$\ds \sum_{k \mathop = 1}^n \dfrac 1 k \sum_{j \mathop = 1}^k \dfrac 1 j$

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 j \dfrac 1 k$

$\ds $

$=$

$\ds \dfrac 1 2 \paren {\paren {\sum_{k \mathop = 1}^n \dfrac 1 k}^2 + \paren {\sum_{k \mathop = 1}^n \dfrac 1 {k^2} } }$

$\ds $

$=$

$\ds \dfrac { {H_n}^2 + \harm 2 n } 2$

Hence the result.

$\blacksquare$