Proving $\sum_{n=1}^{\infty}\dfrac{H^{(a)}_{n}}{n^b} = \zeta(a,b) + \zeta(a+b)$

Question

I've come across the formula:

$$\sum_{n=1}^{\infty}\dfrac{H^{(a)}_{n}}{n^b} = \zeta(a,b) + \zeta(a+b)$$

where $H^{(a)}_{n}$ is the Generalized Harmonic nuber and $\zeta(a,b)$ is the Multiple zeta function.

I have no idea if this identity is true, and if yes then how to prove it? Thanks.

Answer 1

Considering $$f(a,b)=\sum_{n=1}^{\infty}\dfrac{H^{(a)}_{n}}{n^b} \qquad \text{and} \qquad g(a,b)=\zeta(a,b) + \zeta(a+b)$$ I just computed a few values and it does not seem to work $$\left( \begin{array}{cccc} a & b & f(a,b) & g(a,b) \\ 6 & 5 & 1.03751 & 1.00060 \\ 6 & 4 & 1.08364 & 1.00134 \\ 6 & 3 & 1.20533 & 1.00373 \\ 6 & 2 & 1.65564 & 1.02142 \\ 5 & 4 & 1.08499 & 1.00357 \\ 5 & 3 & 1.20874 & 1.00976 \\ 5 & 2 & 1.66710 & 1.04528 \\ 4 & 3 & 1.21585 & 1.02817 \\ 4 & 2 & 1.69187 & 1.09967 \\ 3 & 2 & 1.74849 & 1.23898 \\ \end{array} \right)$$

Answer 2

Based on the link you provided, we have $$\zeta(a,b)=\sum_{n=1}^\infty\frac{H_n^{(a)}}{(n+1)^b}$$

Let the index start from zero since $H_0^{(a)}=0$,

$$\zeta(a,b)=\sum_{n=0}^\infty\frac{H_n^{(a)}}{(n+1)^b}$$

Shift the index,

$$\zeta(a,b)=\sum_{n=1}^\infty\frac{H_{n-1}^{(a)}}{n^b}$$

Use $H_{n-1}^{(a)}=H_n^{(a)}-\frac1{n^a}$,

$$\zeta(a,b)=\sum_{n=1}^\infty\frac{H_n^{(a)}}{n^b}-\sum_{n=1}^\infty\frac{1}{n^{a+b}}=\sum_{n=1}^\infty\frac{H_n^{(a)}}{n^b}-\zeta(a+b)$$

or

$$\sum_{n=1}^{\infty}\dfrac{H^{(a)}_{n}}{n^b} = \zeta(a,b) + \zeta(a+b).$$