Summation to n of Square of kth Harmonic Number

Theorem

 * $\displaystyle \sum_{k \mathop = 1}^n {H_k}^2 = \left({n + 1}\right) {H_n}^2 - \left({2 n + 1}\right) H_n + 2 n$

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

Proof
Consider

Using Summation by Parts: