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?
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 $n=1,..,k-1$).
$\zeta(s) - H_n(s) = \sum_{m=n+1}^{\infty} \dfrac1{m^s}$ Hence, $\begin{align} \sum_{n=1}^{\infty} (\zeta(s) - H_n(s)) & = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \dfrac1{m^s}\\& = \sum_{m=2}^{\infty} \sum_{n=1}^{m-1} \dfrac1{m^s} \text{ (Changing the order of summation)}\\& = \sum_{m=2}^{\infty} \left( \dfrac1{m^{s-1}} - \dfrac1{m^s}\right)\\ & = \sum_{m=1}^{\infty} \left( \dfrac1{m^{s-1}} - \dfrac1{m^s}\right)\\ & = \zeta(s-1) - \zeta(s) \end{align} $