# Closed form expression or asymptotic expansion for (periodic) generalized harmonic numbers?

In contrast with the series $$\sum_{k=1}^n k$$ and $$\sum_{k=1}^n1$$, there does not (as far as I know) exist a pure closed form expression (or a nice asymptotic expansion other than the Euler-Maclaurin expansion?) for the generalized harmonic numbers (and in fact the defining series for the Riemann zeta function) $$H_{s}(n)=\sum_{k=1}^n \frac{1}{k^s}\tag{1}$$ with $$s\in\mathbb C$$ (in particular $$\Re(s)=\sigma\in(0,1)$$).

My question is however whether there might exist a closed form expression (or maybe an asymptotic expansion other than the Euler-Maclaurin expansion?) of which I call 'periodic generalized harmonic numbers' $$P_{\sigma,t}(n)$$ and $$Q_{\sigma,t}(n)$$ (however an expression for one of them would suffice) in which $$P_{\sigma,t}(n)=\sum_{k=1}^n \frac{\cos(t\ln(k))}{k^\sigma}\tag{2a}$$ and $$Q_{\sigma,t}(n)=\sum_{k=1}^n \frac{\sin(t\ln(k))}{k^\sigma}\tag{2b}$$ with $$t\in\mathbb R$$ and (in particular) $$\sigma\in(0,1)$$.

The reason why I'm not immediately interested in the Euler-Maclaurin expansion of the three concerning series is that all these expansions involve a complicated expression (involving Bernoulli numbers and derivatives of the concerning terms) which is in fact used to define the Riemann zeta function in some way. I'm in particular looking for a method to derive information from the Riemann zeta function by using another type of asymptotic expansion or even the closed form of the defining series (1) or preferably the (maybe more nicely behaving) 'splitted versions' (2a) and (2b). Is e.g. Fourier analysis an obvious direction to think of?

See for my motivation for this matter also in 4.1.2 of http://fse.studenttheses.ub.rug.nl/19062/1/bMATH_2019_vanderReijdenIS.pdf.

• Welcome to the site. This is a good question. I can't answer it, but I know there are certain users here who can. Hope they see it. – Antonio Vargas Dec 23 '18 at 20:45
• Thank you! I hope so: Without giving full disclosure I've found a method which, when operating on the according partial sums expressions (if they do exist of course), might give us necessary conditions for the real part of a zero of the Riemann zeta function (hence might solve the Riemann hypothesis). My thesis is about this matter but I strongly suppose I can't find a partial sum expression by myself; when one can indeed find such an expression and the according method gives more insights in the real part of the zeroes of the zeta function I will let him/her personally know ;). – Bas van der Reijden Dec 23 '18 at 21:19
• Could you explain what do you mean by "both series are divergent"? The statement cannot mean $P_{n,m,\theta}$, can it? – user Apr 18 '19 at 9:52
• All partial sum expressions are indeed finite hence convergent ;). The series are divergent when $m$ (I've rewrited it to $a$) is smaller than 1 and $n\to\infty$ (I've noted this from the concerning Euler-Maclaurin expansions), this fact is not really important and was more or less written to note that one doesn't have to use 'classical tricks' to determine a convergence value when $n\to\infty$. I've rewritten the question to emphasize the context in more detail and give more direction to it. – Bas van der Reijden Apr 19 '19 at 13:35

Since $$\sum\limits_{k=1}^n \dfrac{e^{ib\ln k}}{k^a} = \sum\limits_{k=1}^n \dfrac{k^{ib}}{k^a} = \sum\limits_{k=1}^n\dfrac1{k^{a-ib}} = H_{a-ib}(n),$$ then $$P_{a,b}(n) = \Re H_{a-ib}(n),\quad Q_{a,b}(n) = \Im H_{a-ib}(n).$$

Is known that $$H_z(n) = \zeta(z) - \zeta(z,n+1),$$ where $$\zeta(z)$$ is the Riemann zeta function and $$\zeta(z,k)$$ is the Hurwitz zeta function.

This allows to choose the most suitable presentation for the goal function.

For example, can be used the integral presentation in the form of $$\zeta(z,n+1) = \dfrac1{\Gamma(z)}\int\limits_0^\infty \dfrac{t^{z-1}}{1-e^{-t}}\,e^{-(n+1)t}\,dt$$ and similarly for Riemann zeta (case $$n=-1$$).

The periodicity of Hurwitz zeta can be used in the representation of $$\zeta(z, n+1) = 2(2\pi)^{z-1}\Gamma(1-z)\left(\sin\frac{\pi z}2\sum\limits_{k=1}^{\infty} \cos 2\pi(n+1)k) + \cos\frac{\pi z}2\sum\limits_{k=1}^{\infty} \sin2\pi(n+1)k)\right)$$ (also look there).

• Thanks! I've seen this identity earlier but I don't see how it is useful in my context? Since the zeta functions are (indirectly) the object of study, I've used its properties to derive the 'periodic parts' (1) and (2). What do you mean by the last sentence? – Bas van der Reijden Apr 19 '19 at 20:16
• @Frozenharp Thanks for the comment, I've detalized this in the answer. – Yuri Negometyanov Apr 19 '19 at 20:42
• Ok, this integral representation looks useful! But as far as I know it is only valid when $\Re(z)>1$?... And (correct me if I'm wrong) I think you might have interchanged $z$ and $n+1$ in the integral expression?? – Bas van der Reijden Apr 19 '19 at 21:29
• @Frozenharp You are right. Thanks, fixed. Is it useful now? – Yuri Negometyanov Apr 19 '19 at 22:02
• I'm afraid the 'periodic formula' for the Hurwitz-zeta function will only be valid when $0<n+1\leq1$ hence not very useful in my case?? Nevertheless, I think some other representation of the Hurwitz zeta function might be useful and I will do some research on it the upcoming time. – Bas van der Reijden Apr 22 '19 at 14:27