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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?

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up vote 5 down vote accepted

$$\sum_{n=1}^{\infty} (\zeta(s) - H_n(s)) = \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{m^s}$$ $$= \sum_{k=2}^{\infty} \frac{k-1}{k^s} = \sum_{k=2}^{\infty} k^{1-s} - \sum_{k=2}^{\infty} k^{-s} $$$$=\zeta(s-1) - 1 - (\zeta(s) - 1) = \zeta(s-1) - \zeta(s)$$

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 $n=1,..,k-1$).

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And the interchange $\sum_{n=1}^\infty \sum_{m=n+1}^\infty = \sum_{m=2}^\infty \sum_{n=1}^{m-1}$ is justified by absolute convergence for $\text{Re}(s-1) > 1$. – Robert Israel Jun 24 '12 at 21:15
Thanks, Cocopuffs. It is good to know it is true. – Tito Piezas III Jun 24 '12 at 21:24

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

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Thanks, Marvis. It seems MS does not allow 2 answers to be accepted, so I chose the first one. Yours is elegant too. – Tito Piezas III Jun 24 '12 at 21:29
@TitoPiezasIII No worries. Both of us answered roughly at the same time and Cocopuffs answer has an earlier time stamp. So it is perfectly fine accepting his answer. – user17762 Jun 24 '12 at 21:31

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