Summation to n of Square of kth Harmonic Number

Theorem

 * $\displaystyle \sum_{k \mathop = 1}^n {H_k}^2 = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n$

where $H_k$ denotes the $k$th harmonic number.

Proof
Consider:

Using Summation by Parts: