Summation to n of kth Harmonic Number over k+1

Theorem

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

where:
 * $H_n$ denotes the $n$th harmonic number
 * $H_n^{\paren 2}$ denotes a general harmonic number.

Proof
Hence the result.