Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers 2

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

share|improve this answer
3  
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
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.