Riemann zeta sums and harmonic numbers Given the nth harmonic number of order s,
$$H_n(s) =\sum_{m=1}^n \frac{1}{m^s}$$
It can be empirically observed that, for $s > 2$, then,
$$\sum_{n=1}^\infty\Big[\zeta(s)-H_n(s)\Big] = \zeta(s-1)-\zeta(s)$$
Can anyone prove this is true?

$$\sum_{n=1}^{\infty} (\zeta(s) - H_n(s)) = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{m^s}$$ $$= \sum_{k=2}^{\infty} \frac{k-1}{k^s} = \sum_{k=2}^{\infty} k^{1-s} - \sum_{k=2}^{\infty} k^{-s} $$$$=\zeta(s-1) - 1 - (\zeta(s) - 1) =  \zeta(s-1) - \zeta(s)$$
since each term $\frac{1}{k^s}$ appears in exactly $k-1$ of the sums $\sum_{m=n+1}^{\infty} \frac{1}{m^2}$ (namely, for $n=1,..,k-1$).
