I want to show the following equation
$\sum_{k=1}^n \frac{1}{k} H_k = \frac{1}{2} (H_n^2 + H_n^{(2)})$
where $H_k$ is the k-th harmonic number and $H_k^{(2)}$ is the second generalized harmonic number.
I tried summation by parts
$\sum_{k=1}^n \frac{1}{k} H_k = H_n \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^{n-1} (H_{k+1}- H_k) \sum_{i=1}^k \frac{1}{i} = H_n^2 - \sum_{k=1}^{n-1} \frac{1}{k+1} H_k $
but now I've no idea how to bring $H_k^{(2)}$ in the equation. Can someone please give me a hint?
Observe that $$\sum_{k=1}^n\frac{H_k}k=\sum_{1\le j\le k\le n}\frac1{jk},$$ $$H_n^2=\sum_{1\le j, k\le n}\frac1{jk},$$ $$H_n^{(2)}=\sum_{1\le j=k\le n}\frac1{jk}$$ etc.