# Intruiging Symmetric harmonic sum $\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$

I proved the following equation

$$\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$$

We define

$$H^{(k)}_n=\sum_{m= 1}^n \frac{1}{m^k}$$

I am looking forward to seeing what approaches would you use .

#### Solutions Collecting From Web of "Intruiging Symmetric harmonic sum $\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$"

Isn’t this just
\begin{align} \zeta^2(k) &= \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{n^km^k} \\&= 2\sum_{1\leq m\leq n} \frac{1}{(nm)^k} – \sum_{m=1}^\infty\frac{1}{(n^k)^2}\\ &=2\sum_{n=1}^\infty \frac{1}{n^k}\sum_{m=1}^n \frac{1}{m^k} – \zeta(2k)\\ &=2\sum_{n=1}^\infty \frac{H_n^{(k)}}{n^k} – \zeta(2k) \end{align}?

You could also figure that:

$$\zeta^2(k) = 2\sum_{n=1}^\infty \frac{H_{n-1}^{(k)}}{n^k} +\zeta(2k)$$
Every rearrangement here can be done due to absolute convergence.