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
Using Summation by Parts:


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

Then:

and:

Thus:

So: